भाग 1 — समाधान (Solutions)
राजस्थान बोर्ड कक्षा 10 गणित — सभी अभ्यास एवं प्रश्नावली के चरण-दर-चरण सटीक हल।
#### प्रश्नावली 8.1
प्रश्न [1]. Δ A B C \Delta ABC Δ A B C में, जिसका कोण B B B समकोण है, A B = 24 c m AB=24 cm A B = 24 c m और B C = 7 c m BC=7 cm B C = 7 c m है। निम्नलिखित का मान ज्ञात कीजिए :
(i) s i n A , c o s A sinA,cosA s in A , cos A (ii) s i n C , c o s C sinC,cosC s in C , cos C
हल— (i) ज्ञात करना है s i n A , c o s A sinA,cosA s in A , cos A
यहाँ A B = 24 c m AB=24 cm A B = 24 c m B C = 7 c m BC=7 cm B C = 7 c m
पाइथागोरस प्रमेय से
A C 2 = A B 2 + B C 2 AC^{2}=AB^{2}+BC^{2} A C 2 = A B 2 + B C 2
A C 2 = ( 24 ) 2 + ( 7 ) 2 AC^{2}=(24)^{2}+(7)^{2} A C 2 = ( 24 ) 2 + ( 7 ) 2
A C 2 = 576 + 49 AC^{2}=576+49 A C 2 = 576 + 49
[Right-angled triangle ABC with right angle at B, AB=24cm, BC=7cm, AC=25cm]
A C 2 = 625 AC^{2}=625 A C 2 = 625
A C = 625 AC=\sqrt{625} A C = 625
A C = 25 c m AC=25 cm A C = 25 c m
∵ s i n A = B C A C ∵sinA=\frac{BC}{AC} ∵ s in A = A C B C
∴ s i n A = 7 c m 25 c m = 7 25 ∴sinA=\frac{7 cm}{25 cm}=\frac{7}{25} ∴ s in A = 25 c m 7 c m = 25 7
तथा c o s A = A B A C = 24 c m 25 c m cosA=\frac{AB}{AC}=\frac{24 cm}{25 cm} cos A = A C A B = 25 c m 24 c m
या c o s A = 24 25 cosA=\frac{24}{25} cos A = 25 24
(ii) s i n C = ∠ C कीसम्मुखभुजा कर्ण sinC=\frac{ \angle C \text{की} \text{सम्मुख} \text{भुजा}}{\text{कर्ण}} s in C = कर्ण ∠ C की सम्मुख भुजा
= A B A C = 24 c m 25 c m =\frac{AB}{AC}=\frac{24 cm}{25 cm} = A C A B = 25 c m 24 c m
∴ s i n C = 24 25 ∴sinC=\frac{24}{25} ∴ s in C = 25 24
c o s C = ∠ C कीआधारभुजा कर्ण cosC=\frac{ \angle C \text{की} \text{आधार} \text{भुजा}}{\text{कर्ण}} cos C = कर्ण ∠ C की आधार भुजा
= B C A C = 7 c m 25 c m =\frac{BC}{AC}=\frac{7 cm}{25 cm} = A C B C = 25 c m 7 c m
∴ c o s C = 7 25 ∴cosC=\frac{7}{25} ∴ cos C = 25 7
**2. आकृति में, t a n P − c o t R tanP-cotR t an P − co tR
[समकोण त्रिभुज PQR जिसमें ∠ Q = 90°, PQ = 12 cm, PR = 13 cm]
हल— कर्ण P R = 13 c m PR=13 cm P R = 13 c m
पाइथागोरस प्रमेय से
P R 2 = P Q 2 + Q R 2 PR^{2}=PQ^{2}+QR^{2} P R 2 = P Q 2 + Q R 2
या ( 13 ) 2 = ( 12 ) 2 + ( Q R ) 2 (13)^{2}=(12)^{2}+(QR)^{2} ( 13 ) 2 = ( 12 ) 2 + ( QR ) 2
या 169 = 144 + ( Q R ) 2 169=144+(QR)^{2} 169 = 144 + ( QR ) 2
या 169 − 144 = ( Q R ) 2 169-144=(QR)^{2} 169 − 144 = ( QR ) 2
या 25 = ( Q R ) 2 25=(QR)^{2} 25 = ( QR ) 2
या Q R = ± 25 QR= \pm \sqrt{25} QR = ± 25
या Q R = 5 , − 5 QR=5,-5 QR = 5 , − 5
परन्तु Q R = 5 c m QR=5 cm QR = 5 c m
( Q R ≠ − 5 ) (QR \ne -5) ( QR = − 5 ) ] ] ]
[समकोण त्रिभुज PQR]
t a n P = ∠ P कीसम्मुखभुजा ∠ P कीआधारभुजा tanP=\frac{ \angle P \text{की} \text{सम्मुख} \text{भुजा}}{ \angle P \text{की} \text{आधार} \text{भुजा}} t an P = ∠ P की आधार भुजा ∠ P की सम्मुख भुजा
∴ t a n P = R Q Q P = 5 12 ∴tanP=\frac{RQ}{QP}=\frac{5}{12} ∴ t an P = QP R Q = 12 5
और c o t R = ∠ R कीआधारभुजा ∠ R कीसम्मुखभुजा cotR=\frac{ \angle R \text{की} \text{आधार} \text{भुजा}}{ \angle R \text{की} \text{सम्मुख} \text{भुजा}} co tR = ∠ R की सम्मुख भुजा ∠ R की आधार भुजा
∴ c o t R = R Q P Q = 5 12 ∴cotR=\frac{RQ}{PQ}=\frac{5}{12} ∴ co tR = P Q R Q = 12 5
तब t a n P − c o t R = 5 12 − 5 12 = 0 tanP-cotR=\frac{5}{12}-\frac{5}{12}=0 t an P − co tR = 12 5 − 12 5 = 0 उत्तर
**3. यदि s i n A = 3 4 sinA=\frac{3}{4} s in A = 4 3 c o s A cosA cos A t a n A tanA t an A
( माध्य. शिक्षा बोर्ड, मॉडल पेपर, 2021-22 )
हल— माना कि A B C ABC A B C कोई समकोण त्रिभुज है जिसमें कोण B B B पर समकोण है।
s i n A = 3 4 sinA=\frac{3}{4} s in A = 4 3
परन्तु s i n A = लम्ब कर्ण = B C A C sinA=\frac{\text{लम्ब}}{\text{कर्ण}}=\frac{BC}{AC} s in A = कर्ण लम्ब = A C B C
[समकोण त्रिभुज ABC जिसमें ∠ B = 90°, AC = 4K, BC = 3K, AB = √(7)K]
∴ B C A C = 3 4 ∴\frac{BC}{AC}=\frac{3}{4} ∴ A C B C = 4 3
माना B C = 3 K BC=3K B C = 3 K
A C = 4 K AC=4K A C = 4 K
पाइथागोरस प्रमेय से
A C 2 = A B 2 + B C 2 AC^{2}=AB^{2}+BC^{2} A C 2 = A B 2 + B C 2
या ( 4 K ) 2 = ( A B ) 2 + ( 3 K ) 2 (4K)^{2}=(AB)^{2}+(3K)^{2} ( 4 K ) 2 = ( A B ) 2 + ( 3 K ) 2
या 16 K 2 = A B 2 + 9 K 2 16K^{2}=AB^{2}+9K^{2} 16 K 2 = A B 2 + 9 K 2
या 16 K 2 − 9 K 2 = A B 2 16K^{2}-9K^{2}=AB^{2} 16 K 2 − 9 K 2 = A B 2
या 7 K 2 = A B 2 7K^{2}=AB^{2} 7 K 2 = A B 2
या A B = ± 7 K 2 AB= \pm \sqrt{7K^{2}} A B = ± 7 K 2
या A B = ± 7 K AB= \pm \sqrt{7}K A B = ± 7 K
( A B ≠ − 7 K ) (AB \ne -\sqrt{7}K) ( A B = − 7 K ) ] ] ]
⇒ A B = 7 K ⇒AB=\sqrt{7}K ⇒ A B = 7 K
∴ c o s A = आधार कर्ण = A B A C ∴cosA=\frac{\text{आधार}}{\text{कर्ण}}=\frac{AB}{AC} ∴ cos A = कर्ण आधार = A C A B
c o s A = 7 K 4 K = 7 4 cosA=\frac{\sqrt{7}K}{4K}=\frac{\sqrt{7}}{4} cos A = 4 K 7 K = 4 7 उत्तर
तथा t a n A = लम्ब आधार = B C A B = 3 K 7 K = 3 7 tanA=\frac{\text{लम्ब}}{\text{आधार}}=\frac{BC}{AB}=\frac{3K}{\sqrt{7}K}=\frac{3}{\sqrt{7}} t an A = आधार लम्ब = A B B C = 7 K 3 K = 7 3 उत्तर
**4. यदि 15 c o t A = 8 15cotA=8 15 co t A = 8 s i n A sinA s in A s e c A secA sec A
हल— माना कि A B C ABC A B C कोई समकोण त्रिभुज है जिसमें A A A न्यून कोण है और B B B पर समकोण है।
[समकोण त्रिभुज ABC जिसमें ∠ B = 90°, AC = 17K, BC = 15K, AB = 8K]
15 c o t A = 8 15cotA=8 15 co t A = 8
c o t A = 8 15 cotA=\frac{8}{15} co t A = 15 8
परन्तु c o t A = आधार लम्ब = A B B C cotA=\frac{\text{आधार}}{\text{लम्ब}}=\frac{AB}{BC} co t A = लम्ब आधार = B C A B
⇒ A B B C = 8 15 ⇒\frac{AB}{BC}=\frac{8}{15} ⇒ B C A B = 15 8
माना A B = 8 K , B C = 15 K AB=8K,BC=15K A B = 8 K , B C = 15 K
पाइथागोरस प्रमेय से
A C 2 = ( A B ) 2 + ( B C ) 2 AC^{2}=(AB)^{2}+(BC)^{2} A C 2 = ( A B ) 2 + ( B C ) 2
( A C ) 2 = ( 8 K ) 2 + ( 15 K ) 2 (AC)^{2}=(8K)^{2}+(15K)^{2} ( A C ) 2 = ( 8 K ) 2 + ( 15 K ) 2
( A C ) 2 = 64 K 2 + 225 K 2 ) (AC)^{2}=64K^{2}+225K^{2}) ( A C ) 2 = 64 K 2 + 225 K 2 )
( A C ) 2 ) (AC)^{2}) ( A C ) 2 )
A C = ± 289 K 2 AC= \pm \sqrt{289K^{2}} A C = ± 289 K 2
A C = ± 17 K AC= \pm 17K A C = ± 17 K
⇒ A C = 17 K ⇒AC=17K ⇒ A C = 17 K
( A C = − 17 K , क्योंकिभुजाऋणात्मकनहींहोसकती ) (AC=-17K,\text{क्योंकि} \text{भुजा} \text{ऋणात्मक} \text{नहीं} \text{हो} \text{सकती}) ( A C = − 17 K , क्योंकि भुजा ऋणात्मक नहीं हो सकती )
s i n A = लम्ब कर्ण = B C A C = 15 K 17 K = 15 17 sinA=\frac{\text{लम्ब}}{\text{कर्ण}}=\frac{BC}{AC}=\frac{15K}{17K}=\frac{15}{17} s in A = कर्ण लम्ब = A C B C = 17 K 15 K = 17 15
और s e c A = कर्ण आधार = A C A B secA=\frac{\text{कर्ण}}{\text{आधार}}=\frac{AC}{AB} sec A = आधार कर्ण = A B A C
s e c A = 17 K 8 K = 17 8 secA=\frac{17K}{8K}=\frac{17}{8} sec A = 8 K 17 K = 8 17
**5. यदि s e c θ = 13 12 sec \theta =\frac{13}{12} sec θ = 12 13
हल—माना कि A B C ABC A B C कोई समकोण त्रिभुज है जिसमें B B B पर समकोण है।
[Right-angled triangle ABC with angle θ at A, side AB=12k, BC=5k, AC=13k]
पुन: माना कि ∠ B A C = θ \angle BAC= \theta ∠ B A C = θ
प्रश्नानुसार
s e c θ = 13 12 = कर्ण आधार = A C A B sec \theta =\frac{13}{12}=\frac{\text{कर्ण}}{\text{आधार}}=\frac{AC}{AB} sec θ = 12 13 = आधार कर्ण = A B A C
∴ s e c θ = A C A B ∴sec \theta =\frac{AC}{AB} ∴ sec θ = A B A C
A C A B = 13 12 \frac{AC}{AB}=\frac{13}{12} A B A C = 12 13
माना A C = 13 k AC=13k A C = 13 k A B = 12 k AB=12k A B = 12 k
पाइथागोरस प्रमेय से
A C 2 = ( A B ) 2 + ( B C ) 2 AC^{2}=(AB)^{2}+(BC)^{2} A C 2 = ( A B ) 2 + ( B C ) 2
या ( 13 k ) 2 = ( 12 k ) 2 + ( B C ) 2 (13k)^{2}=(12k)^{2}+(BC)^{2} ( 13 k ) 2 = ( 12 k ) 2 + ( B C ) 2
या 169 k 2 = 144 k 2 + ( B C ) 2 169k^{2}=144k^{2}+(BC)^{2} 169 k 2 = 144 k 2 + ( B C ) 2
या 169 k 2 − 144 k 2 = ( B C ) 2 169k^{2}-144k^{2}=(BC)^{2} 169 k 2 − 144 k 2 = ( B C ) 2
या ( B C ) 2 ) (BC)^{2}) ( B C ) 2 )
या B C = ± 25 k 2 BC= \pm \sqrt{25k^{2}} B C = ± 25 k 2
या B C = ± 5 k BC= \pm 5k B C = ± 5 k
या B C = 5 k BC=5k B C = 5 k
( B C ≠ − 5 k क्योंकिभुजाऋणात्मकनहींहोसकती ) (BC \ne -5k \text{क्योंकि} \text{भुजा} \text{ऋणात्मक} \text{नहीं} \text{हो} \text{सकती}) ( B C = − 5 k क्योंकि भुजा ऋणात्मक नहीं हो सकती )
s i n θ = लम्ब कर्ण = B C A C = 5 k 13 k = 5 13 sin \theta =\frac{\text{लम्ब}}{\text{कर्ण}}=\frac{BC}{AC}=\frac{5k}{13k}=\frac{5}{13} s in θ = कर्ण लम्ब = A C B C = 13 k 5 k = 13 5
c o s θ = आधार कर्ण = A B A C = 12 k 13 k = 12 13 cos \theta =\frac{\text{आधार}}{\text{कर्ण}}=\frac{AB}{AC}=\frac{12k}{13k}=\frac{12}{13} cos θ = कर्ण आधार = A C A B = 13 k 12 k = 13 12
t a n θ = लम्ब आधार = B C A B = 5 k 12 k = 5 12 tan \theta =\frac{\text{लम्ब}}{\text{आधार}}=\frac{BC}{AB}=\frac{5k}{12k}=\frac{5}{12} t an θ = आधार लम्ब = A B B C = 12 k 5 k = 12 5
c o s e c θ = कर्ण लम्ब = A C B C = 13 k 5 k = 13 5 cosec \theta =\frac{\text{कर्ण}}{\text{लम्ब}}=\frac{AC}{BC}=\frac{13k}{5k}=\frac{13}{5} cosec θ = लम्ब कर्ण = B C A C = 5 k 13 k = 5 13
c o t θ = आधार लम्ब = A B B C = 12 k 5 k = 12 5 cot \theta =\frac{\text{आधार}}{\text{लम्ब}}=\frac{AB}{BC}=\frac{12k}{5k}=\frac{12}{5} co tθ = लम्ब आधार = B C A B = 5 k 12 k = 5 12
**6. यदि ∠ A \angle A ∠ A ∠ B \angle B ∠ B c o s A = c o s B cosA=cosB cos A = cos B ∠ A = ∠ B \angle A= \angle B ∠ A = ∠ B
हल—माना कि A B C ABC A B C कोई त्रिभुज है जहाँ ∠ A \angle A ∠ A और ∠ B \angle B ∠ B न्यून कोण हैं। c o s A cosA cos A और c o s B cosB cos B ज्ञात करने हैं।
अब C M ⊥ A B CM⟂AB C M ⊥ A B
[Triangle ABC with altitude CM from C to AB]
∴ ∠ A M C = ∠ B M C = 90 ∘ ∴ \angle AMC= \angle BMC=90^{∘} ∴ ∠ A M C = ∠ B M C = 9 0 ∘
समकोण Δ A M C \Delta AMC Δ A M C
A M A C = c o s A \frac{AM}{AC}=cosA A C A M = cos A
पुन: समकोण Δ B M C \Delta BMC Δ B M C
B M B C = c o s B \frac{BM}{BC}=cosB B C B M = cos B
परन्तु c o s A = c o s B cosA=cosB cos A = cos B
(i), (ii) और (iii) से,
A M A C = B M B C \frac{AM}{AC}=\frac{BM}{BC} A C A M = B C B M
A M B M = A C B C = C M C M \frac{AM}{BM}=\frac{AC}{BC}=\frac{CM}{CM} B M A M = B C A C = C M C M
∴ Δ A M C Δ B M C ∴ \Delta AMC~ \Delta BMC ∴ Δ A M C Δ B M C
⇒ ∠ A = ∠ B ⇒ \angle A= \angle B ⇒ ∠ A = ∠ B
( ) () ( )
**7. यदि c o t θ = 7 8 cot \theta =\frac{7}{8} co tθ = 8 7
**(i) ( 1 + s i n θ ) ( 1 − s i n θ ) ( 1 + c o s θ ) ( 1 − c o s θ ) \frac{(1+sin \theta )(1-sin \theta )}{(1+cos \theta )(1-cos \theta )} ( 1 + cos θ ) ( 1 − cos θ ) ( 1 + s in θ ) ( 1 − s in θ )
(ii) c o t 2 θ cot^{2} \theta co t 2 θ
हल—(i) ∠ A B C = θ \angle ABC= \theta ∠ A B C = θ
समकोण त्रिभुज A C B ACB A C B C C C
[Right-angled triangle ABC with angle θ at B, sides BC=7k, AC=8k, AB=√(113)k]
प्रश्नानुसार
c o t θ = 7 k 8 k = 7 8 cot \theta =\frac{7k}{8k}=\frac{7}{8} co tθ = 8 k 7 k = 8 7
परन्तु c o t θ = B C A C cot \theta =\frac{BC}{AC} co tθ = A C B C
⇒ B C A C = 7 8 ⇒\frac{BC}{AC}=\frac{7}{8} ⇒ A C B C = 8 7
माना B C = 7 k , A C = 8 k BC=7k,AC=8k B C = 7 k , A C = 8 k
पाइथागोरस प्रमेय से
A B 2 = ( B C ) 2 + ( A C ) 2 AB^{2}=(BC)^{2}+(AC)^{2} A B 2 = ( B C ) 2 + ( A C ) 2
या ( A B ) 2 = ( 7 k ) 2 + ( 8 k ) 2 (AB)^{2}=(7k)^{2}+(8k)^{2} ( A B ) 2 = ( 7 k ) 2 + ( 8 k ) 2
या ( A B ) 2 = 49 k 2 + 64 k 2 ) (AB)^{2}=49k^{2}+64k^{2}) ( A B ) 2 = 49 k 2 + 64 k 2 )
या ( A B ) 2 ) (AB)^{2}) ( A B ) 2 )
या A B = ± 113 k 2 AB= \pm \sqrt{113k^{2}} A B = ± 113 k 2
या A B = 113 k AB=\sqrt{113}k A B = 113 k
[ A B ≠ − 113 k AB \ne -\sqrt{113}k A B = − 113 k
∴ s i n θ = A C A B = 8 k 113 k ∴sin \theta =\frac{AC}{AB}=\frac{8k}{\sqrt{113}k} ∴ s in θ = A B A C = 113 k 8 k
⇒ s i n θ = 8 113 ⇒sin \theta =\frac{8}{\sqrt{113}} ⇒ s in θ = 113 8
c o s θ = B C A B = 7 k 113 k = 7 113 cos \theta =\frac{BC}{AB}=\frac{7k}{\sqrt{113}k}=\frac{7}{\sqrt{113}} cos θ = A B B C = 113 k 7 k = 113 7
c o s θ = 7 113 cos \theta =\frac{7}{\sqrt{113}} cos θ = 113 7
( 1 + s i n θ ) ( 1 − s i n θ ) (1+sin \theta )(1-sin \theta ) ( 1 + s in θ ) ( 1 − s in θ )
= 1 + 8 113 1 − 8 113 =1+\frac{8}{\sqrt{113}}1-\frac{8}{\sqrt{113}} = 1 + 113 8 1 − 113 8
= ( 1 ) 2 − 8 113 2 =(1)^{2}-\frac{8}{\sqrt{113}}^{2} = ( 1 ) 2 − 113 8 2
[सूत्र ( a + b ) ( a − b ) = a 2 − b 2 (a+b)(a-b)=a^{2}-b^{2} ( a + b ) ( a − b ) = a 2 − b 2
= 1 − 64 113 =1-\frac{64}{113} = 1 − 113 64
= 113 − 64 113 = 49 113 =\frac{113-64}{113}=\frac{49}{113} = 113 113 − 64 = 113 49
⇒ ( 1 + s i n θ ) ( 1 − s i n θ ) = 49 113 ⇒(1+sin \theta )(1-sin \theta )=\frac{49}{113} ⇒ ( 1 + s in θ ) ( 1 − s in θ ) = 113 49
अब ( 1 + c o s θ ) ( 1 − c o s θ ) (1+cos \theta )(1-cos \theta ) ( 1 + cos θ ) ( 1 − cos θ )
= 1 + 7 113 1 − 7 113 =1+\frac{7}{\sqrt{113}}1-\frac{7}{\sqrt{113}} = 1 + 113 7 1 − 113 7
= ( 1 ) 2 − 7 113 2 =(1)^{2}-\frac{7}{\sqrt{113}}^{2} = ( 1 ) 2 − 113 7 2
[सूत्र ( a + b ) ( a − b ) = a 2 − b 2 (a+b)(a-b)=a^{2}-b^{2} ( a + b ) ( a − b ) = a 2 − b 2
= 1 − 49 113 =1-\frac{49}{113} = 1 − 113 49
= 113 − 49 113 =\frac{113-49}{113} = 113 113 − 49
∴ ( 1 + c o s θ ) ( 1 − c o s θ ) = 64 113 ∴(1+cos \theta )(1-cos \theta )=\frac{64}{113} ∴ ( 1 + cos θ ) ( 1 − cos θ ) = 113 64
∴ ( 1 + s i n θ ) ( 1 − s i n θ ) ( 1 + c o s θ ) ( 1 − c o s θ ) = 49 113 64 113 ∴\frac{(1+sin \theta )(1-sin \theta )}{(1+cos \theta )(1-cos \theta )}=\frac{\frac{49}{113}}{\frac{64}{113}} ∴ ( 1 + cos θ ) ( 1 − cos θ ) ( 1 + s in θ ) ( 1 − s in θ ) = 113 64 113 49
अतः:
( 1 + s i n θ ) ( 1 − s i n θ ) ( 1 + c o s θ ) ( 1 − c o s θ ) = 49 64 \frac{(1+sin \theta )(1-sin \theta )}{(1+cos \theta )(1-cos \theta )}=\frac{49}{64} ( 1 + cos θ ) ( 1 − cos θ ) ( 1 + s in θ ) ( 1 − s in θ ) = 64 49
(ii) c o t θ = B C A C = 7 8 cot \theta =\frac{BC}{AC}=\frac{7}{8} co tθ = A C B C = 8 7
c o t 2 θ = ( c o t θ ) 2 cot^{2} \theta =(cot \theta )^{2} co t 2 θ = ( co tθ ) 2
c o t 2 θ = 7 8 2 cot^{2} \theta =\frac{7}{8}^{2} co t 2 θ = 8 7 2
⇒ c o t 2 θ = 49 64 ⇒cot^{2} \theta =\frac{49}{64} ⇒ co t 2 θ = 64 49
अतः: c o t 2 θ cot^{2} \theta co t 2 θ = 49 64 =\frac{49}{64} = 64 49
प्रश्न 8. यदि 3 c o t A = 4 3cotA=4 3 co t A = 4 तो जाँच कीजिए कि 1 − t a n 2 A 1 + t a n 2 A = c o s 2 A − s i n 2 A \frac{1-tan^{2}A}{1+tan^{2}A}=cos^{2}A-sin^{2}A 1 + t a n 2 A 1 − t a n 2 A = co s 2 A − s i n 2 A है या नहीं।
हल—माना कि A B C ABC A B C एक समकोण त्रिभुज है जिसमें B B B पर समकोण है।
प्रश्नानुसार 3 c o t A = 4 3cotA=4 3 co t A = 4
∴ c o t A = 4 3 ∴cotA=\frac{4}{3} ∴ co t A = 3 4
परन्तु c o t A = A B B C cotA=\frac{AB}{BC} co t A = B C A B
⇒ A B B C = 4 3 ⇒\frac{AB}{BC}=\frac{4}{3} ⇒ B C A B = 3 4
माना A B = 4 k , B C = 3 k AB=4k,BC=3k A B = 4 k , B C = 3 k
[Right-angled triangle ABC with angle A at A, sides AB=4k, BC=3k, AC=5k]
पाइथागोरस प्रमेय से
( A C ) 2 = ( A B ) 2 + ( B C ) 2 (AC)^{2}=(AB)^{2}+(BC)^{2} ( A C ) 2 = ( A B ) 2 + ( B C ) 2
( A C ) 2 = ( 4 k ) 2 + ( 3 k ) 2 (AC)^{2}=(4k)^{2}+(3k)^{2} ( A C ) 2 = ( 4 k ) 2 + ( 3 k ) 2
( A C ) 2 = 16 k 2 + 9 k 2 ) (AC)^{2}=16k^{2}+9k^{2}) ( A C ) 2 = 16 k 2 + 9 k 2 )
( A C ) 2 ) (AC)^{2}) ( A C ) 2 )
A C = ± 25 k 2 AC= \pm \sqrt{25k^{2}} A C = ± 25 k 2
A C = ± 5 k AC= \pm 5k A C = ± 5 k
परन्तु A C = 5 k AC=5k A C = 5 k
( A C ≠ − 5 k ) (AC \ne -5k) ( A C = − 5 k ) ] ] ]
s i n A = B C A C = 3 k 5 k = 3 5 sinA=\frac{BC}{AC}=\frac{3k}{5k}=\frac{3}{5} s in A = A C B C = 5 k 3 k = 5 3
t a n A = B C A B = 3 k 4 k = 3 4 tanA=\frac{BC}{AB}=\frac{3k}{4k}=\frac{3}{4} t an A = A B B C = 4 k 3 k = 4 3
c o s A = A B A C = 4 k 5 k = 4 5 cosA=\frac{AB}{AC}=\frac{4k}{5k}=\frac{4}{5} cos A = A C A B = 5 k 4 k = 5 4
L . H . S . = 1 − t a n 2 A 1 + t a n 2 A L.H.S.=\frac{1-tan^{2}A}{1+tan^{2}A} L . H . S . = 1 + t a n 2 A 1 − t a n 2 A
= 1 − ( 3 4 ) 2 1 + ( 3 4 ) 2 ∵ t a n A = 3 4 =\frac{1-(\frac{3}{4})^{2}}{1+(\frac{3}{4})^{2}}∵tanA=\frac{3}{4} = 1 + ( 4 3 ) 2 1 − ( 4 3 ) 2 ∵ t an A = 4 3
= 1 − 9 16 1 + 9 16 = 16 − 9 16 16 + 9 16 =\frac{1-\frac{9}{16}}{1+\frac{9}{16}}=\frac{\frac{16-9}{16}}{\frac{16+9}{16}} = 1 + 16 9 1 − 16 9 = 16 16 + 9 16 16 − 9
= 7 16 25 16 = 7 25 =\frac{\frac{7}{16}}{\frac{25}{16}}=\frac{7}{25} = 16 25 16 7 = 25 7
∴ 1 − t a n 2 A 1 + t a n 2 A = 7 25 ∴\frac{1-tan^{2}A}{1+tan^{2}A}=\frac{7}{25} ∴ 1 + t a n 2 A 1 − t a n 2 A = 25 7
R . H . S . = c o s 2 A − s i n 2 A R.H.S.=cos^{2}A-sin^{2}A R . H . S . = co s 2 A − s i n 2 A
= ( 4 5 ) 2 − ( 3 5 ) 2 =(\frac{4}{5})^{2}-(\frac{3}{5})^{2} = ( 5 4 ) 2 − ( 5 3 ) 2
= 16 25 − 9 25 =\frac{16}{25}-\frac{9}{25} = 25 16 − 25 9
= 16 − 9 25 = 7 25 =\frac{16-9}{25}=\frac{7}{25} = 25 16 − 9 = 25 7
∴ c o s 2 A − s i n 2 A = 7 25 ∴cos^{2}A-sin^{2}A=\frac{7}{25} ∴ co s 2 A − s i n 2 A = 25 7
(i) और (ii) से,
L . H . S . = R . H . S . L.H.S.=R.H.S. L . H . S . = R . H . S .
∴ 1 − t a n 2 A 1 + t a n 2 A = c o s 2 A − s i n 2 A ∴\frac{1-tan^{2}A}{1+tan^{2}A}=cos^{2}A-sin^{2}A ∴ 1 + t a n 2 A 1 − t a n 2 A = co s 2 A − s i n 2 A
**9. त्रिभुज ABC में, जिसका कोण B समकोण है, यदि t a n A = 1 3 tanA=\frac{1}{\sqrt{3}} t an A = 3 1
(i) s i n A c o s C + c o s A s i n C sinAcosC+cosAsinC s in A cos C + cos A s in C
(ii) c o s A c o s C − s i n A s i n C cosAcosC-sinAsinC cos A cos C − s in A s in C
हल— (i) प्रश्नानुसार Δ A B C \Delta ABC Δ A B C जिसका कोण B समकोण है।
[Right angled triangle ABC with angle B = 90 degrees, AB = √(3)k, BC = k, AC = 2k]
t a n A = 1 3 tanA=\frac{1}{\sqrt{3}} t an A = 3 1
परन्तु t a n A = B C A B tanA=\frac{BC}{AB} t an A = A B B C
(i) और (ii) से,
B C A B = 1 3 \frac{BC}{AB}=\frac{1}{\sqrt{3}} A B B C = 3 1
माना B C = k , A B = 3 k BC=k,AB=\sqrt{3}k B C = k , A B = 3 k
पाइथागोरस प्रमेय से,
( A C ) 2 = ( A B ) 2 + ( B C ) 2 (AC)^{2}=(AB)^{2}+(BC)^{2} ( A C ) 2 = ( A B ) 2 + ( B C ) 2
या ( A C ) 2 = ( 3 k ) 2 + ( k ) 2 (AC)^{2}=(\sqrt{3}k)^{2}+(k)^{2} ( A C ) 2 = ( 3 k ) 2 + ( k ) 2
या A C 2 = 3 k 2 + k 2 AC^{2}=3k^{2}+k^{2} A C 2 = 3 k 2 + k 2
या A C 2 = 4 k 2 AC^{2}=4k^{2} A C 2 = 4 k 2
या A C = ± 4 k 2 AC= \pm \sqrt{4k^{2}} A C = ± 4 k 2
A C = ± 2 k AC= \pm 2k A C = ± 2 k
जहाँ A C = 2 k AC=2k A C = 2 k
( A C ≠ − 2 k ) (AC \ne -2k) ( A C = − 2 k ) ] ] ]
s i n A = B C A C = k 2 k = 1 2 sinA=\frac{BC}{AC}=\frac{k}{2k}=\frac{1}{2} s in A = A C B C = 2 k k = 2 1
c o s C = B C A C = k 2 k = 1 2 cosC=\frac{BC}{AC}=\frac{k}{2k}=\frac{1}{2} cos C = A C B C = 2 k k = 2 1
c o s A = A B A C = 3 k 2 k = 3 2 cosA=\frac{AB}{AC}=\frac{\sqrt{3}k}{2k}=\frac{\sqrt{3}}{2} cos A = A C A B = 2 k 3 k = 2 3
s i n C = A B A C = 3 k 2 k = 3 2 sinC=\frac{AB}{AC}=\frac{\sqrt{3}k}{2k}=\frac{\sqrt{3}}{2} s in C = A C A B = 2 k 3 k = 2 3
s i n A c o s C = ( 1 2 ) ( 1 2 ) = 1 4 sinAcosC=(\frac{1}{2})(\frac{1}{2})=\frac{1}{4} s in A cos C = ( 2 1 ) ( 2 1 ) = 4 1
c o s A s i n C = ( 3 2 ) ( 3 2 ) = 3 4 cosAsinC=(\frac{\sqrt{3}}{2})(\frac{\sqrt{3}}{2})=\frac{3}{4} cos A s in C = ( 2 3 ) ( 2 3 ) = 4 3
∴ s i n A c o s C + c o s A s i n C ∴sinAcosC+cosAsinC ∴ s in A cos C + cos A s in C
= 1 4 + 3 4 =\frac{1}{4}+\frac{3}{4} = 4 1 + 4 3
= 1 + 3 4 = 4 4 = 1 =\frac{1+3}{4}=\frac{4}{4}=1 = 4 1 + 3 = 4 4 = 1
(ii) c o s A c o s C = ( 3 2 ) ( 1 2 ) = 3 4 cosAcosC=(\frac{\sqrt{3}}{2})(\frac{1}{2})=\frac{\sqrt{3}}{4} cos A cos C = ( 2 3 ) ( 2 1 ) = 4 3
s i n A s i n C = ( 1 2 ) ( 3 2 ) = 3 4 sinAsinC=(\frac{1}{2})(\frac{\sqrt{3}}{2})=\frac{\sqrt{3}}{4} s in A s in C = ( 2 1 ) ( 2 3 ) = 4 3
∴ c o s A c o s C − s i n A s i n C ∴cosAcosC-sinAsinC ∴ cos A cos C − s in A s in C
= ( 3 4 ) − ( 3 4 ) = 0 =(\frac{\sqrt{3}}{4})-(\frac{\sqrt{3}}{4})=0 = ( 4 3 ) − ( 4 3 ) = 0
**10. Δ P Q R \Delta PQR Δ P QR Q Q Q P R + Q R = 25 c m PR+QR=25 cm P R + QR = 25 c m P Q = 5 c m PQ=5 cm P Q = 5 c m s i n P , c o s P sinP,cosP s in P , cos P t a n P tanP t an P
हल—प्रश्नानुसार Δ P Q R \Delta PQR Δ P QR , में Q Q Q पर समकोण है।
P R + Q R = 25 c m PR+QR=25 cm P R + QR = 25 c m
P Q = 5 c m PQ=5 cm P Q = 5 c m
समकोण त्रिभुज P Q R PQR P QR
[Right angled triangle PQR with right angle at Q]
पाइथागोरस प्रमेय से
( P R ) 2 = ( P Q ) 2 + ( R Q ) 2 (PR)^{2}=(PQ)^{2}+(RQ)^{2} ( P R ) 2 = ( P Q ) 2 + ( R Q ) 2
या ( P R ) 2 = ( 5 ) 2 + ( R Q ) 2 (PR)^{2}=(5)^{2}+(RQ)^{2} ( P R ) 2 = ( 5 ) 2 + ( R Q ) 2
∵ P R + Q R = 25 ∵PR+QR=25 ∵ P R + QR = 25
∴ Q R = 25 − P R ∴QR=25-PR ∴ QR = 25 − P R
या ( P R ) 2 = 25 + [ 25 − P R ] 2 ) (PR)^{2}=25+[25-PR]^{2}) ( P R ) 2 = 25 + [ 25 − P R ] 2 )
या ( P R ) 2 = 25 + ( 25 ) 2 + ( P R ) 2 − 2 × 25 × P R ) (PR)^{2}=25+(25)^{2}+(PR)^{2}-2 \times 25 \times PR) ( P R ) 2 = 25 + ( 25 ) 2 + ( P R ) 2 − 2 × 25 × P R )
या ( P R ) 2 = 25 + 625 + ( P R ) 2 − 50 P R ) (PR)^{2}=25+625+(PR)^{2}-50PR) ( P R ) 2 = 25 + 625 + ( P R ) 2 − 50 P R )
या ( P R ) 2 − ( P R ) 2 + 50 P R = 650 ) (PR)^{2}-(PR)^{2}+50PR=650) ( P R ) 2 − ( P R ) 2 + 50 P R = 650 )
या 50 P R = 650 50PR=650 50 P R = 650
या P R = 650 50 PR=\frac{650}{50} P R = 50 650
या P R = 13 PR=13 P R = 13
Q R = 25 − P R QR=25-PR QR = 25 − P R
Q R = 25 − 13 QR=25-13 QR = 25 − 13
या Q R = 12 c m QR=12 cm QR = 12 c m
s i n P = Q R P R = 12 13 sinP=\frac{QR}{PR}=\frac{12}{13} s in P = P R QR = 13 12
c o s P = P Q P R = 5 13 cosP=\frac{PQ}{PR}=\frac{5}{13} cos P = P R P Q = 13 5
t a n P = Q R P Q = 12 5 tanP=\frac{QR}{PQ}=\frac{12}{5} t an P = P Q QR = 5 12
11. बताइए कि निम्नलिखित कथन सत्य हैं या असत्य। कारण सहित अपने उत्तर की पुष्टि कीजिए :
(i) t a n A tanA t an A 1 1 1
(ii) कोण A A A s e c A = 12 5 secA=\frac{12}{5} sec A = 5 12
(iii) c o s A cosA cos A A A A c o s e c a n t cosecant cosec an t
(iv) c o t A , c o t cotA,cot co t A , co t A A A
(v) किसी भी कोण θ \theta θ s i n θ = 4 3 sin \theta =\frac{4}{3} s in θ = 3 4
हल—(i) असत्य है।
चूँकि समकोण त्रिभुज की भुजाओं की लम्बाई के मान कुछ भी हो सकते हैं। आधार की लम्बाई लम्ब से छोटी भी हो सकती है और बड़ी भी हो सकती है। अतः t a n A tanA t an A 1 1 1 1 1 1
(ii) s e c A = कर्ण आधार = 12 5 = 2.40 > 1 secA=\frac{\text{कर्ण}}{\text{आधार}}=\frac{12}{5}=2.40>1 sec A = आधार कर्ण = 5 12 = 2.40 > 1
∵ ∵ ∵ 12 : 5 12:5 12 : 5
ऐसा सदैव होना आवश्यक नहीं है, पर किसी मान के लिए सत्य है।
अतः कथन सत्य है।
(iii) ∵ c o s A ∵cosA ∵ cos A A A A c o s i n e cosine cos in e c o s e c a n t A cosecant A cosec an t A c o s e c A cosec A cosec A
(iv) असत्य।
क्योंकि c o t A cotA co t A A A A c o t a n g e n t cotangent co t an g e n t c o t cot co t A A A
(v) असत्य s i n θ = 4 3 = 1.666 > 1 sin \theta =\frac{4}{3}=1.666>1 s in θ = 3 4 = 1.666 > 1
क्योंकि s i n θ sin \theta s in θ 1 1 1 1 1 1
1. निम्नलिखित के मान निकालिए :
(i) s i n 60 ∘ c o s 30 ∘ + s i n 30 ∘ c o s 60 ∘ sin60^{∘}cos30^{∘}+sin30^{∘}cos60^{∘} s in 6 0 ∘ cos 3 0 ∘ + s in 3 0 ∘ cos 6 0 ∘
(ii) 2 t a n 2 45 ∘ + c o s 2 30 ∘ − s i n 2 60 ∘ 2tan^{2}45^{∘}+cos^{2}30^{∘}-sin^{2}60^{∘} 2 t a n 2 4 5 ∘ + co s 2 3 0 ∘ − s i n 2 6 0 ∘
(iii) c o s 45 ∘ s e c 30 ∘ + c o s e c 30 ∘ \frac{cos45^{∘}}{sec30^{∘}+cosec 30^{∘}} sec 3 0 ∘ + cosec 3 0 ∘ cos 4 5 ∘
(iv) s i n 30 ∘ + t a n 45 ∘ − c o s e c 60 ∘ s e c 30 ∘ + c o s 60 ∘ + c o t 45 ∘ \frac{sin30^{∘}+tan45^{∘}-cosec 60^{∘}}{sec30^{∘}+cos60^{∘}+cot45^{∘}} sec 3 0 ∘ + cos 6 0 ∘ + co t 4 5 ∘ s in 3 0 ∘ + t an 4 5 ∘ − cosec 6 0 ∘
(v) 5 c o s 2 60 ∘ + 4 s e c 2 30 ∘ − t a n 2 45 ∘ s i n 2 30 ∘ + c o s 2 30 ∘ \frac{5cos^{2}60^{∘}+4sec^{2}30^{∘}-tan^{2}45^{∘}}{sin^{2}30^{∘}+cos^{2}30^{∘}} s i n 2 3 0 ∘ + co s 2 3 0 ∘ 5 co s 2 6 0 ∘ + 4 se c 2 3 0 ∘ − t a n 2 4 5 ∘
हल—
(i) प्रश्नानुसार
s i n 60 ∘ c o s 30 ∘ + s i n 30 ∘ c o s 60 ∘ sin60^{∘}cos30^{∘}+sin30^{∘}cos60^{∘} s in 6 0 ∘ cos 3 0 ∘ + s in 3 0 ∘ cos 6 0 ∘
= ( 3 2 ) ( 3 2 ) + ( 1 2 ) ( 1 2 ) =(\frac{\sqrt{3}}{2})(\frac{\sqrt{3}}{2})+(\frac{1}{2})(\frac{1}{2}) = ( 2 3 ) ( 2 3 ) + ( 2 1 ) ( 2 1 )
= ( 3 2 ) 2 + ( 1 2 ) 2 =(\frac{\sqrt{3}}{2})^{2}+(\frac{1}{2})^{2} = ( 2 3 ) 2 + ( 2 1 ) 2
= 3 4 + 1 4 = 1 =\frac{3}{4}+\frac{1}{4}=1 = 4 3 + 4 1 = 1 उत्तर
(ii) 2 t a n 2 45 ∘ + c o s 2 30 ∘ − s i n 2 60 ∘ 2tan^{2}45^{∘}+cos^{2}30^{∘}-sin^{2}60^{∘} 2 t a n 2 4 5 ∘ + co s 2 3 0 ∘ − s i n 2 6 0 ∘
= 2 ( t a n 45 ∘ ) 2 + ( c o s 30 ∘ ) 2 − ( s i n 60 ∘ ) 2 =2(tan45^{∘})^{2}+(cos30^{∘})^{2}-(sin60^{∘})^{2} = 2 ( t an 4 5 ∘ ) 2 + ( cos 3 0 ∘ ) 2 − ( s in 6 0 ∘ ) 2
= 2 ( 1 ) 2 + ( 3 2 ) 2 − ( 3 2 ) 2 =2(1)^{2}+(\frac{\sqrt{3}}{2})^{2}-(\frac{\sqrt{3}}{2})^{2} = 2 ( 1 ) 2 + ( 2 3 ) 2 − ( 2 3 ) 2
= 2 × 1 + 3 4 − 3 4 = 2 + 3 4 − 3 4 =2 \times 1+\frac{3}{4}-\frac{3}{4}=2+\frac{3}{4}-\frac{3}{4} = 2 × 1 + 4 3 − 4 3 = 2 + 4 3 − 4 3
= 2 =2 = 2 उत्तर
(iii) c o s 45 ∘ s e c 30 ∘ + c o s e c 30 ∘ \frac{cos45^{∘}}{sec30^{∘}+cosec 30^{∘}} sec 3 0 ∘ + cosec 3 0 ∘ cos 4 5 ∘
= 1 2 2 3 + ( 2 ) = 1 2 2 + 2 3 3 =\frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}}+(2)}=\frac{\frac{1}{\sqrt{2}}}{\frac{2+2\sqrt{3}}{\sqrt{3}}} = 3 2 + ( 2 ) 2 1 = 3 2 + 2 3 2 1
= 1 2 × 3 2 + 2 3 = 3 2 2 ( 3 + 1 ) =\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2+2\sqrt{3}}=\frac{\sqrt{3}}{2\sqrt{2}(\sqrt{3}+1)} = 2 1 × 2 + 2 3 3 = 2 2 ( 3 + 1 ) 3
= 3 ( 3 − 1 ) 2 2 ( 3 + 1 ) ( 3 − 1 ) =\frac{\sqrt{3}(\sqrt{3}-1)}{2\sqrt{2}(\sqrt{3}+1)(\sqrt{3}-1)} = 2 2 ( 3 + 1 ) ( 3 − 1 ) 3 ( 3 − 1 ) ( 3 ) (\sqrt{3}) ( 3 )
= 3 ( 3 − 1 ) 2 2 ( 3 − 1 ) =\frac{\sqrt{3}(\sqrt{3}-1)}{2\sqrt{2}(3-1)} = 2 2 ( 3 − 1 ) 3 ( 3 − 1 )
= 2 × 3 × ( 3 − 1 ) 2 × 2 2 ( 3 − 1 ) =\frac{\sqrt{2} \times \sqrt{3} \times (\sqrt{3}-1)}{\sqrt{2} \times 2\sqrt{2}(3-1)} = 2 × 2 2 ( 3 − 1 ) 2 × 3 × ( 3 − 1 ) 2 \sqrt{2} 2
= 2 × 3 × ( 3 − 1 ) 4 ( 3 − 1 ) = 3 2 − 6 8 =\frac{\sqrt{2} \times \sqrt{3} \times (\sqrt{3}-1)}{4(3-1)}=\frac{3\sqrt{2}-\sqrt{6}}{8} = 4 ( 3 − 1 ) 2 × 3 × ( 3 − 1 ) = 8 3 2 − 6 उत्तर
(iv) s i n 30 ∘ + t a n 45 ∘ − c o s e c 60 ∘ s e c 30 ∘ + c o s 60 ∘ + c o t 45 ∘ \frac{sin30^{∘}+tan45^{∘}-cosec 60^{∘}}{sec30^{∘}+cos60^{∘}+cot45^{∘}} sec 3 0 ∘ + cos 6 0 ∘ + co t 4 5 ∘ s in 3 0 ∘ + t an 4 5 ∘ − cosec 6 0 ∘
= 1 2 + 1 − 2 3 2 3 + 1 2 + 1 = 3 2 − 2 3 2 3 + 3 2 = 3 3 − 4 2 3 4 + 3 3 2 3 =\frac{\frac{1}{2}+1-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{1}{2}+1}=\frac{\frac{3}{2}-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{3}{2}}=\frac{\frac{3\sqrt{3}-4}{2\sqrt{3}}}{\frac{4+3\sqrt{3}}{2\sqrt{3}}} = 3 2 + 2 1 + 1 2 1 + 1 − 3 2 = 3 2 + 2 3 2 3 − 3 2 = 2 3 4 + 3 3 2 3 3 3 − 4
= 3 3 − 4 4 + 3 3 =\frac{3\sqrt{3}-4}{4+3\sqrt{3}} = 4 + 3 3 3 3 − 4
= ( 3 3 − 4 ) ( 3 3 − 4 ) ( 3 3 + 4 ) ( 3 3 − 4 ) =\frac{(3\sqrt{3}-4)(3\sqrt{3}-4)}{(3\sqrt{3}+4)(3\sqrt{3}-4)} = ( 3 3 + 4 ) ( 3 3 − 4 ) ( 3 3 − 4 ) ( 3 3 − 4 )
= ( 3 3 − 4 ) 2 ( 3 3 ) 2 − ( 4 ) 2 =\frac{(3\sqrt{3}-4)^{2}}{(3\sqrt{3})^{2}-(4)^{2}} = ( 3 3 ) 2 − ( 4 ) 2 ( 3 3 − 4 ) 2
= ( 3 3 ) 2 + ( 4 ) 2 − 2 × 3 3 × 4 ) 9 × 3 − 16 =\frac{(3\sqrt{3})^{2}+(4)^{2}-2 \times 3\sqrt{3} \times 4)}{9 \times 3-16} = 9 × 3 − 16 ( 3 3 ) 2 + ( 4 ) 2 − 2 × 3 3 × 4 )
= 27 + 16 − 24 3 27 − 16 =\frac{27+16-24\sqrt{3}}{27-16} = 27 − 16 27 + 16 − 24 3
= 43 − 24 3 11 =\frac{43-24\sqrt{3}}{11} = 11 43 − 24 3 उत्तर
(v) 5 c o s 2 60 ∘ + 4 s e c 2 30 ∘ − t a n 2 45 ∘ s i n 2 30 ∘ + c o s 2 30 ∘ \frac{5cos^{2}60^{∘}+4sec^{2}30^{∘}-tan^{2}45^{∘}}{sin^{2}30^{∘}+cos^{2}30^{∘}} s i n 2 3 0 ∘ + co s 2 3 0 ∘ 5 co s 2 6 0 ∘ + 4 se c 2 3 0 ∘ − t a n 2 4 5 ∘
= 5 ( c o s 60 ∘ ) 2 + 4 ( s e c 30 ∘ ) 2 − ( t a n 45 ∘ ) 2 ( s i n 30 ∘ ) 2 + ( c o s 30 ∘ ) 2 =\frac{5(cos60^{∘})^{2}+4(sec30^{∘})^{2}-(tan45^{∘})^{2}}{(sin30^{∘})^{2}+(cos30^{∘})^{2}} = ( s in 3 0 ∘ ) 2 + ( cos 3 0 ∘ ) 2 5 ( cos 6 0 ∘ ) 2 + 4 ( sec 3 0 ∘ ) 2 − ( t an 4 5 ∘ ) 2
= 5 ( 1 2 ) 2 + 4 ( 2 3 ) 2 − ( 1 ) 2 ( 1 2 ) 2 + ( 3 2 ) 2 =\frac{5(\frac{1}{2})^{2}+4(\frac{2}{\sqrt{3}})^{2}-(1)^{2}}{(\frac{1}{2})^{2}+(\frac{\sqrt{3}}{2})^{2}} = ( 2 1 ) 2 + ( 2 3 ) 2 5 ( 2 1 ) 2 + 4 ( 3 2 ) 2 − ( 1 ) 2
= 5 4 + 4 × 4 3 − 1 1 4 + 3 4 = 5 4 + 16 3 − 1 1 4 + 3 4 =\frac{\frac{5}{4}+4 \times \frac{4}{3}-1}{\frac{1}{4}+\frac{3}{4}}=\frac{\frac{5}{4}+\frac{16}{3}-1}{\frac{1}{4}+\frac{3}{4}} = 4 1 + 4 3 4 5 + 4 × 3 4 − 1 = 4 1 + 4 3 4 5 + 3 16 − 1
= 5 4 + 16 3 − 1 = 15 + 64 − 12 12 =\frac{5}{4}+\frac{16}{3}-1=\frac{15+64-12}{12} = 4 5 + 3 16 − 1 = 12 15 + 64 − 12
= 67 12 =\frac{67}{12} = 12 67
2. सही विकल्प चुनिए और अपने विकल्प का औचित्य दीजिए :
(i) 2 t a n 30 ∘ 1 + t a n 2 30 ∘ = \frac{2tan30^{∘}}{1+tan^{2}30^{∘}}= 1 + t a n 2 3 0 ∘ 2 t an 3 0 ∘ =
(A) s i n 60 ∘ sin60^{∘} s in 6 0 ∘ (B) c o s 60 ∘ cos60^{∘} cos 6 0 ∘ (C) t a n 60 ∘ tan60^{∘} t an 6 0 ∘ (D) s i n 30 ∘ sin30^{∘} s in 3 0 ∘
(ii) 1 − t a n 2 45 ∘ 1 + t a n 2 45 ∘ = \frac{1-tan^{2}45^{∘}}{1+tan^{2}45^{∘}}= 1 + t a n 2 4 5 ∘ 1 − t a n 2 4 5 ∘ =
(A) t a n 90 ∘ tan90^{∘} t an 9 0 ∘ (C) s i n 45 ∘ sin45^{∘} s in 4 5 ∘
(iii) s i n 2 A = 2 s i n A sin2A=2sinA s in 2 A = 2 s in A A A A
(iv) 2 t a n 30 ∘ 1 − t a n 2 30 ∘ \frac{2tan30^{∘}}{1-tan^{2}30^{∘}} 1 − t a n 2 3 0 ∘ 2 t an 3 0 ∘
(A) c o s 60 ∘ cos60^{∘} cos 6 0 ∘ (B) s i n 60 ∘ sin60^{∘} s in 6 0 ∘ (C) t a n 60 ∘ tan60^{∘} t an 6 0 ∘ (D) s i n 30 ∘ sin30^{∘} s in 3 0 ∘
हल—(i) 2 t a n 30 ∘ 1 + t a n 2 30 ∘ = 2 ( 1 3 ) 1 + ( 1 3 ) 2 \frac{2tan30^{∘}}{1+tan^{2}30^{∘}}=\frac{2(\frac{1}{\sqrt{3}})}{1+(\frac{1}{\sqrt{3}})^{2}} 1 + t a n 2 3 0 ∘ 2 t an 3 0 ∘ = 1 + ( 3 1 ) 2 2 ( 3 1 )
= 2 3 1 + 1 3 = 2 3 × 3 4 = 3 2 = s i n 60 ∘ =\frac{\frac{2}{\sqrt{3}}}{1+\frac{1}{3}}=\frac{2}{\sqrt{3}} \times \frac{3}{4}=\frac{\sqrt{3}}{2}=sin60^{∘} = 1 + 3 1 3 2 = 3 2 × 4 3 = 2 3 = s in 6 0 ∘
सही विकल्प = (A) उत्तर
(ii) 1 − t a n 2 45 ∘ 1 + t a n 2 45 ∘ = 1 − ( 1 ) 2 1 + ( 1 ) 2 = 0 \frac{1-tan^{2}45^{∘}}{1+tan^{2}45^{∘}}=\frac{1-(1)^{2}}{1+(1)^{2}}=0 1 + t a n 2 4 5 ∘ 1 − t a n 2 4 5 ∘ = 1 + ( 1 ) 2 1 − ( 1 ) 2 = 0
सही विकल्प = (D) उत्तर
(iii) प्रश्नानुसार s i n 2 A = 2 s i n A sin2A=2sinA s in 2 A = 2 s in A
जब A = 0 ∘ A=0^{∘} A = 0 ∘
s i n 2 ( 0 ) = 2 s i n 0 sin2(0)=2sin0 s in 2 ( 0 ) = 2 s in 0
s i n 0 = 0 sin0=0 s in 0 = 0
0 = 0 0=0 0 = 0
∴ ∴ ∴
(iv) 2 t a n 30 ∘ 1 − t a n 2 30 ∘ = 2 ( 1 3 ) 1 − ( 1 3 ) 2 \frac{2tan30^{∘}}{1-tan^{2}30^{∘}}=\frac{2(\frac{1}{\sqrt{3}})}{1-(\frac{1}{\sqrt{3}})^{2}} 1 − t a n 2 3 0 ∘ 2 t an 3 0 ∘ = 1 − ( 3 1 ) 2 2 ( 3 1 )
= 2 3 1 − 1 3 = 2 3 × 3 2 = 3 = t a n 60 ∘ =\frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}}=\frac{2}{\sqrt{3}} \times \frac{3}{2}=\sqrt{3}=tan60^{∘} = 1 − 3 1 3 2 = 3 2 × 2 3 = 3 = t an 6 0 ∘
सही विकल्प = (C) उत्तर
**3. यदि t a n ( A + B ) = 3 tan(A+B)=\sqrt{3} t an ( A + B ) = 3 t a n ( A − B ) = 1 3 tan(A-B)=\frac{1}{\sqrt{3}} t an ( A − B ) = 3 1 0 ∘ B 0^{∘} B 0 ∘ B A A A B B B
( माध्य. शिक्षा बोर्ड, मॉडल पेपर, 2024-25 )
हल— t a n ( A + B ) = 3 tan(A+B)=\sqrt{3} t an ( A + B ) = 3
t a n ( A + B ) = t a n 60 ∘ tan(A+B)=tan60^{∘} t an ( A + B ) = t an 6 0 ∘
या A + B = 60 ∘ A+B=60^{∘} A + B = 6 0 ∘
t a n ( A − B ) = 1 3 tan(A-B)=\frac{1}{\sqrt{3}} t an ( A − B ) = 3 1
या t a n ( A − B ) = t a n 30 ∘ tan(A-B)=tan30^{∘} t an ( A − B ) = t an 3 0 ∘
या A − B = 30 ∘ A-B=30^{∘} A − B = 3 0 ∘
समीकरण (i) और समीकरण (ii) को जोड़ने पर
A + B = 60 ∘ A+B=60^{∘} A + B = 6 0 ∘
A − B = 30 ∘ A-B=30^{∘} A − B = 3 0 ∘
2 A = 90 ∘ 2A=90^{∘} 2 A = 9 0 ∘
A = 90 ∘ 2 = 45 ∘ A=\frac{90^{∘}}{2}=45^{∘} A = 2 9 0 ∘ = 4 5 ∘
A = 45 ∘ A=45^{∘} A = 4 5 ∘
समीकरण (i) में A A A
45 ∘ + B = 60 ∘ 45^{∘}+B=60^{∘} 4 5 ∘ + B = 6 0 ∘
B = 60 ∘ − 45 ∘ B=60^{∘}-45^{∘} B = 6 0 ∘ − 4 5 ∘
B = 15 ∘ B=15^{∘} B = 1 5 ∘
4. बताइए कि निम्नलिखित में कौन-कौन सत्य हैं या असत्य हैं। कारण सहित अपने उत्तर की पुष्टि कीजिए :
(i) s i n ( A + B ) = s i n A + s i n B sin(A+B)=sinA+sinB s in ( A + B ) = s in A + s in B
(ii) θ \theta θ s i n θ sin \theta s in θ
(iii) θ \theta θ c o s θ cos \theta cos θ
(iv) θ \theta θ s i n θ = c o s θ sin \theta =cos \theta s in θ = cos θ
(v) A = 0 ∘ A=0^{∘} A = 0 ∘ c o t A cotA co t A
हल—
(i) जब A = 60 ∘ , B = 30 ∘ A=60^{∘},B=30^{∘} A = 6 0 ∘ , B = 3 0 ∘
L.H.S. = s i n ( A + B ) =sin(A+B) = s in ( A + B )
= s i n ( 60 ∘ + 30 ∘ ) =sin(60^{∘}+30^{∘}) = s in ( 6 0 ∘ + 3 0 ∘ )
= s i n 90 ∘ = 1 =sin90^{∘}=1 = s in 9 0 ∘ = 1
R.H.S. = s i n A + s i n B =sinA+sinB = s in A + s in B
= s i n 60 ∘ + s i n 30 ∘ =sin60^{∘}+sin30^{∘} = s in 6 0 ∘ + s in 3 0 ∘
= 3 2 + 1 2 ≠ 1 =\frac{\sqrt{3}}{2}+\frac{1}{2} \ne 1 = 2 3 + 2 1 = 1
अर्थात् L.H.S. ≠ \ne =
यह असत्य है। उत्तर
(ii) क्योंकि s i n 0 ∘ = 0 sin0^{∘}=0 s in 0 ∘ = 0
s i n 30 ∘ = 1 2 = 0.5 sin30^{∘}=\frac{1}{2}=0.5 s in 3 0 ∘ = 2 1 = 0.5
s i n 45 ∘ = 1 2 = 0.7 sin45^{∘}=\frac{1}{\sqrt{2}}=0.7 s in 4 5 ∘ = 2 1 = 0.7
s i n 60 ∘ = 3 2 = 0.87 sin60^{∘}=\frac{\sqrt{3}}{2}=0.87 s in 6 0 ∘ = 2 3 = 0.87
और s i n 90 ∘ = 1 sin90^{∘}=1 s in 9 0 ∘ = 1
अर्थात्, जब θ \theta θ 0 ∘ 0^{∘} 0 ∘ 90 ∘ 90^{∘} 9 0 ∘ s i n θ sin \theta s in θ θ = 90 ∘ \theta =90^{∘} θ = 9 0 ∘
∴ ∴ ∴ उत्तर
(iii) क्योंकि c o s 0 ∘ = 1 cos0^{∘}=1 cos 0 ∘ = 1
c o s 30 ∘ = 3 2 = 0.87 cos30^{∘}=\frac{\sqrt{3}}{2}=0.87 cos 3 0 ∘ = 2 3 = 0.87
c o s 45 ∘ = 1 2 = 0.7 cos45^{∘}=\frac{1}{\sqrt{2}}=0.7 cos 4 5 ∘ = 2 1 = 0.7
c o s 60 ∘ = 1 2 = 0.5 cos60^{∘}=\frac{1}{2}=0.5 cos 6 0 ∘ = 2 1 = 0.5
और c o s 90 ∘ = 0 cos90^{∘}=0 cos 9 0 ∘ = 0
जब θ \theta θ 0 ∘ 0^{∘} 0 ∘ 90 ∘ 90^{∘} 9 0 ∘ c o s θ cos \theta cos θ उत्तर
(iv) चूँकि s i n 30 ∘ = 1 2 sin30^{∘}=\frac{1}{2} s in 3 0 ∘ = 2 1
और c o s 30 ∘ = 3 2 cos30^{∘}=\frac{\sqrt{3}}{2} cos 3 0 ∘ = 2 3
या s i n 30 ∘ ≠ c o s 30 ∘ sin30^{∘} \ne cos30^{∘} s in 3 0 ∘ = cos 3 0 ∘
θ = 45 ∘ \theta =45^{∘} θ = 4 5 ∘
s i n 45 ∘ = 1 2 sin45^{∘}=\frac{1}{\sqrt{2}} s in 4 5 ∘ = 2 1
और c o s 45 ∘ = 1 2 cos45^{∘}=\frac{1}{\sqrt{2}} cos 4 5 ∘ = 2 1
सिर्फ θ = 45 ∘ \theta =45^{∘} θ = 4 5 ∘
लेकिन θ \theta θ
s i n θ ≠ c o s θ sin \theta \ne cos \theta s in θ = cos θ
∴ ∴ ∴ उत्तर
(v) c o t 0 ∘ = 1 t a n 0 ∘ = 1 0 cot0^{∘}=\frac{1}{tan0^{∘}}=\frac{1}{0} co t 0 ∘ = t an 0 ∘ 1 = 0 1
यह सत्य है। उत्तर
**1. त्रिकोणमितीय अनुपातों s i n A , s e c A sinA,secA s in A , sec A t a n A tanA t an A c o t A cotA co t A
हल—
(i) ∵ s i n A = 1 c o s e c A ∵sinA=\frac{1}{cosecA} ∵ s in A = cosec A 1
∵ c o s e c 2 A = 1 + c o t 2 A ∵cosec^{2}A=1+cot^{2}A ∵ cose c 2 A = 1 + co t 2 A
∴ c o s e c A = 1 + c o t 2 A ∴cosecA=\sqrt{1+cot^{2}A} ∴ cosec A = 1 + co t 2 A
∴ s i n A = 1 1 + c o t 2 A ∴sinA=\frac{1}{\sqrt{1+cot^{2}A}} ∴ s in A = 1 + co t 2 A 1 उत्तर
(ii) सर्वसमिका का प्रयोग करने पर,
s e c 2 A − t a n 2 A = 1 sec^{2}A-tan^{2}A=1 se c 2 A − t a n 2 A = 1
⇒ s e c 2 A = 1 + t a n 2 A ⇒sec^{2}A=1+tan^{2}A ⇒ se c 2 A = 1 + t a n 2 A
= 1 + 1 c o t 2 A = c o t 2 A + 1 c o t 2 A =1+\frac{1}{cot^{2}A}=\frac{cot^{2}A+1}{cot^{2}A} = 1 + co t 2 A 1 = co t 2 A co t 2 A + 1
⇒ s e c A = c o t 2 A + 1 c o t A ⇒secA=\frac{\sqrt{cot^{2}A+1}}{cotA} ⇒ sec A = co t A co t 2 A + 1 उत्तर
(iii) t a n A = 1 c o t A tanA=\frac{1}{cotA} t an A = co t A 1 उत्तर
**2. ∠ A \angle A ∠ A s e c A secA sec A
गणित-कक्षा 10
181
हल—(i) s i n 2 A + c o s 2 A = 1 sin^{2}A+cos^{2}A=1 s i n 2 A + co s 2 A = 1
⇒ s i n 2 A = 1 − c o s 2 A ⇒sin^{2}A=1-cos^{2}A ⇒ s i n 2 A = 1 − co s 2 A
= 1 − 1 s e c 2 A =1-\frac{1}{sec^{2}A} = 1 − se c 2 A 1
= s e c 2 A − 1 s e c 2 A =\frac{sec^{2}A-1}{sec^{2}A} = se c 2 A se c 2 A − 1
⇒ ( s i n A ) 2 = s e c 2 A − 1 s e c 2 A ⇒(sinA)^{2}=\frac{sec^{2}A-1}{sec^{2}A} ⇒ ( s in A ) 2 = se c 2 A se c 2 A − 1
⇒ s i n A = ± s e c 2 A − 1 s e c A ⇒sinA= \pm \frac{\sqrt{sec^{2}A-1}}{secA} ⇒ s in A = ± sec A se c 2 A − 1
[न्यून कोण A A A
⇒ s i n A = s e c 2 A − 1 s e c A ⇒sinA=\frac{\sqrt{sec^{2}A-1}}{secA} ⇒ s in A = sec A se c 2 A − 1
(ii) c o s A = 1 s e c A cosA=\frac{1}{secA} cos A = sec A 1
(iii) 1 + t a n 2 A = s e c 2 A 1+tan^{2}A=sec^{2}A 1 + t a n 2 A = se c 2 A
t a n 2 A = s e c 2 A − 1 tan^{2}A=sec^{2}A-1 t a n 2 A = se c 2 A − 1
( t a n A ) 2 = s e c 2 A − 1 ) (tanA)^{2}=sec^{2}A-1) ( t an A ) 2 = se c 2 A − 1 )
∴ t a n A = ± s e c 2 A − 1 ∴tanA= \pm \sqrt{sec^{2}A-1} ∴ t an A = ± se c 2 A − 1
[न्यून कोण A A A
अर्थात् t a n A = s e c 2 A − 1 tanA=\sqrt{sec^{2}A-1} t an A = se c 2 A − 1
(iv) c o s e c A = 1 s i n A = 1 s e c 2 A − 1 s e c A cosec A=\frac{1}{sinA}=\frac{1}{\frac{\sqrt{sec^{2}A-1}}{secA}} cosec A = s in A 1 = sec A se c 2 A − 1 1
c o s e c A = s e c A s e c 2 A − 1 cosec A=\frac{secA}{\sqrt{sec^{2}A-1}} cosec A = se c 2 A − 1 sec A
(v) c o t A = 1 t a n A = 1 s e c 2 A − 1 cotA=\frac{1}{tanA}=\frac{1}{\sqrt{sec^{2}A-1}} co t A = t an A 1 = se c 2 A − 1 1
∴ c o t A = 1 s e c 2 A − 1 ∴cotA=\frac{1}{\sqrt{sec^{2}A-1}} ∴ co t A = se c 2 A − 1 1
3. सही विकल्प चुनिए और अपने विकल्प की पुष्टि कीजिए : (प्रश्न बैंक)
(i) 9 s e c 2 A − 9 t a n 2 A 9sec^{2}A-9tan^{2}A 9 se c 2 A − 9 t a n 2 A
(ii) ( 1 + t a n θ + s e c θ ) ( 1 + c o t θ − c o s e c θ ) (1+tan \theta +sec \theta )(1+cot \theta -cosec \theta ) ( 1 + t an θ + sec θ ) ( 1 + co tθ − cosec θ )
(iii) ( s e c A + t a n A ) ( 1 − s i n A ) (secA+tanA)(1-sinA) ( sec A + t an A ) ( 1 − s in A )
(iv) 1 + t a n 2 A 1 + c o t 2 A \frac{1+tan^{2}A}{1+cot^{2}A} 1 + co t 2 A 1 + t a n 2 A
हल—(i) 9 s e c 2 A − 9 t a n 2 A 9sec^{2}A-9tan^{2}A 9 se c 2 A − 9 t a n 2 A
= 9 ( s e c 2 A − t a n 2 A ) =9(sec^{2}A-tan^{2}A) = 9 ( se c 2 A − t a n 2 A )
= 9 × 1 = 9 =9 \times 1=9 = 9 × 1 = 9
∴ ∴ ∴
(ii) ( 1 + t a n θ + s e c θ ) ( 1 + c o t θ − c o s e c θ ) (1+tan \theta +sec \theta )(1+cot \theta -cosec \theta ) ( 1 + t an θ + sec θ ) ( 1 + co tθ − cosec θ )
= 1 + s i n θ c o s θ + 1 c o s θ × 1 + c o s θ s i n θ − 1 s i n θ ={1+\frac{sin \theta }{cos \theta }+\frac{1}{cos \theta }} \times {1+\frac{cos \theta }{sin \theta }-\frac{1}{sin \theta }} = 1 + cos θ s in θ + cos θ 1 × 1 + s in θ cos θ − s in θ 1
= c o s θ + s i n θ + 1 c o s θ × s i n θ + c o s θ − 1 s i n θ ={\frac{cos \theta +sin \theta +1}{cos \theta }} \times {\frac{sin \theta +cos \theta -1}{sin \theta }} = cos θ cos θ + s in θ + 1 × s in θ s in θ + cos θ − 1
= ( ( c o s θ + s i n θ ) + 1 × ( c o s θ + s i n θ ) − 1 ) c o s θ × s i n θ =\frac{((cos \theta +sin \theta )+1} \times {(cos \theta +sin \theta )-1)}{cos \theta \times sin \theta } = × (( cos θ + s in θ ) + 1 ( cos θ + s in θ ) − 1 ) cos θ × s in θ
= ( c o s θ + s i n θ ) 2 − ( 1 ) 2 c o s θ × s i n θ =\frac{(cos \theta +sin \theta )^{2}-(1)^{2}}{cos \theta \times sin \theta } = cos θ × s in θ ( cos θ + s in θ ) 2 − ( 1 ) 2
( ∵ ( a + b ) ( a − b ) = a 2 − b 2 ) (∵(a+b)(a-b)=a^{2}-b^{2}) ( ∵ ( a + b ) ( a − b ) = a 2 − b 2 )
= c o s 2 θ + s i n 2 θ + 2 c o s θ s i n θ − 1 c o s θ × s i n θ =\frac{cos^{2} \theta +sin^{2} \theta +2cos \theta sin \theta -1}{cos \theta \times sin \theta } = cos θ × s in θ co s 2 θ + s i n 2 θ + 2 cos θ s in θ − 1
= 1 + 2 c o s θ s i n θ − 1 c o s θ s i n θ = 2 =\frac{1+2cos \theta sin \theta -1}{cos \theta sin \theta }=2 = cos θ s in θ 1 + 2 cos θ s in θ − 1 = 2
∴ ∴ ∴
(iii) ( s e c A + t a n A ) ( 1 − s i n A ) (secA+tanA)(1-sinA) ( sec A + t an A ) ( 1 − s in A )
= 1 c o s A + s i n A c o s A × ( 1 − s i n A ) =\frac{1}{cosA}+\frac{sinA}{cosA} \times (1-sinA) = cos A 1 + cos A s in A × ( 1 − s in A )
= ( 1 ) c o s A × ( 1 − s i n A ) =\frac{(1)}{cosA} \times (1-sinA) = cos A ( 1 ) × ( 1 − s in A )
= ( 1 + s i n A ) ( 1 − s i n A ) c o s A =\frac{(1+sinA)(1-sinA)}{cosA} = cos A ( 1 + s in A ) ( 1 − s in A )
= ( 1 ) 2 − ( s i n A ) 2 c o s A = 1 − s i n 2 A c o s A = c o s 2 A c o s A =\frac{(1)^{2}-(sinA)^{2}}{cosA}=\frac{1-sin^{2}A}{cosA}=\frac{cos^{2}A}{cosA} = cos A ( 1 ) 2 − ( s in A ) 2 = cos A 1 − s i n 2 A = cos A co s 2 A
( ∵ c o s 2 A = 1 − s i n 2 A ) (∵cos^{2}A=1-sin^{2}A) ( ∵ co s 2 A = 1 − s i n 2 A )
= c o s A =cosA = cos A
∴ ∴ ∴
(iv) 1 + t a n 2 A 1 + c o t 2 A \frac{1+tan^{2}A}{1+cot^{2}A} 1 + co t 2 A 1 + t a n 2 A
= ( s e c 2 A − t a n 2 A ) + t a n 2 A ( c o s e c 2 A − c o t 2 A ) + c o t 2 A =\frac{(sec^{2}A-tan^{2}A)+tan^{2}A}{(cosec^{2}A-cot^{2}A)+cot^{2}A} = ( cose c 2 A − co t 2 A ) + co t 2 A ( se c 2 A − t a n 2 A ) + t a n 2 A
= s e c 2 A c o s e c 2 A = 1 c o s 2 A 1 s i n 2 A =\frac{sec^{2}A}{cosec^{2}A}=\frac{\frac{1}{cos^{2}A}}{\frac{1}{sin^{2}A}} = cose c 2 A se c 2 A = s i n 2 A 1 co s 2 A 1
= s i n 2 A c o s 2 A = s i n A c o s A 2 = t a n 2 A =\frac{sin^{2}A}{cos^{2}A}=\frac{sinA}{cosA}^{2}=tan^{2}A = co s 2 A s i n 2 A = cos A s in A 2 = t a n 2 A
∴ ∴ ∴
4. निम्नलिखित सर्वसमिकाएँ सिद्ध कीजिए जहाँ वे कोण, जिनके लिए व्यंजक परिभाषित है, न्यून कोण हैं :
(i) ( c o s e c θ − c o t θ ) 2 = 1 − c o s θ 1 + c o s θ ) (cosec \theta -cot \theta )^{2}=\frac{1-cos \theta }{1+cos \theta }) ( cosec θ − co tθ ) 2 = 1 + cos θ 1 − cos θ )
( माध्य. शिक्षा बोर्ड, 2024 )
(ii) c o s A 1 + s i n A + 1 + s i n A c o s A = 2 s e c A \frac{cosA}{1+sinA}+\frac{1+sinA}{cosA}=2secA 1 + s in A cos A + cos A 1 + s in A = 2 sec A
( प्रश्न बैंक; माध्य. शिक्षा बोर्ड, मॉडल पेपर, 2023-24, 2022-23 )
(iii) t a n θ 1 − c o t θ + c o t θ 1 − t a n θ = 1 + s e c θ c o s e c θ \frac{tan \theta }{1-cot \theta }+\frac{cot \theta }{1-tan \theta }=1+sec \theta cosec \theta 1 − co tθ t an θ + 1 − t an θ co tθ = 1 + sec θ cosec θ
[ संकेत : व्यंजक को s i n θ sin \theta s in θ c o s θ cos \theta cos θ
(iv) 1 + s e c A s e c A = s i n 2 A 1 − c o s A \frac{1+secA}{secA}=\frac{sin^{2}A}{1-cosA} sec A 1 + sec A = 1 − cos A s i n 2 A
[ संकेत : वाम पक्ष और दायाँ पक्ष को अलग-अलग सरल कीजिए। ]
(v) सर्वसमिका c o s e c 2 A = 1 + c o t 2 A cosec^{2}A=1+cot^{2}A cose c 2 A = 1 + co t 2 A
c o s A − s i n A + 1 c o s A + s i n A − 1 = c o s e c A + c o t A \frac{cosA-sinA+1}{cosA+sinA-1}=cosec A+cotA cos A + s in A − 1 cos A − s in A + 1 = cosec A + co t A
(vi) 1 + s i n A 1 − s i n A = s e c A + t a n A \sqrt{\frac{1+sinA}{1-sinA}}=secA+tanA 1 − s in A 1 + s in A = sec A + t an A
( माध्य. शिक्षा बोर्ड, 2022; माध्य. शिक्षा बोर्ड, मॉडल पेपर, 2022-23 )
(vii) s i n θ − 2 s i n 3 θ 2 c o s 3 θ − c o s θ = t a n θ \frac{sin \theta -2sin^{3} \theta }{2cos^{3} \theta -cos \theta }=tan \theta 2 co s 3 θ − cos θ s in θ − 2 s i n 3 θ = t an θ
( प्रश्न बैंक; माध्य. शिक्षा बोर्ड, मॉडल पेपर, 2023-24 )
(viii) ( s i n A + c o s e c A ) 2 + ( c o s A + s e c A ) 2 = 7 + t a n 2 A + c o t 2 A ) (sinA+cosec A)^{2}+(cosA+secA)^{2}=7+tan^{2}A+cot^{2}A) ( s in A + cosec A ) 2 + ( cos A + sec A ) 2 = 7 + t a n 2 A + co t 2 A )
( माध्य. शिक्षा बोर्ड, 2023 )
(ix) ( c o s e c A − s i n A ) ( s e c A − c o s A ) = 1 t a n A + c o t A (cosec A-sinA)(secA-cosA)=\frac{1}{tanA+cotA} ( cosec A − s in A ) ( sec A − cos A ) = t an A + co t A 1
[ संकेत : वाम पक्ष और दायाँ पक्ष को अलग-अलग सरल कीजिए ]
(x) 1 + t a n 2 A 1 + c o t 2 A = 1 − t a n A 1 − c o t A 2 = t a n 2 A \frac{1+tan^{2}A}{1+cot^{2}A}=\frac{1-tanA}{1-cotA}^{2}=tan^{2}A 1 + co t 2 A 1 + t a n 2 A = 1 − co t A 1 − t an A 2 = t a n 2 A
हल—
(i) L.H.S. = ( c o s e c θ − c o t θ ) 2 (cosec \theta -cot \theta )^{2} ( cosec θ − co tθ ) 2
= 1 s i n θ − c o s θ s i n θ 2 =\frac{1}{sin \theta }-\frac{cos \theta }{sin \theta }^{2} = s in θ 1 − s in θ cos θ 2
= 1 − c o s θ s i n θ 2 = ( 1 − c o s θ ) 2 s i n 2 θ =\frac{1-cos \theta }{sin \theta }^{2}=\frac{(1-cos \theta )^{2}}{sin^{2} \theta } = s in θ 1 − cos θ 2 = s i n 2 θ ( 1 − cos θ ) 2
सर्वसमिका s i n 2 θ + c o s 2 θ = 1 sin^{2} \theta +cos^{2} \theta =1 s i n 2 θ + co s 2 θ = 1
⇒ s i n 2 θ = 1 − c o s 2 θ ⇒sin^{2} \theta =1-cos^{2} \theta ⇒ s i n 2 θ = 1 − co s 2 θ
= ( 1 − c o s θ ) 2 1 − c o s 2 θ =\frac{(1-cos \theta )^{2}}{1-cos^{2} \theta } = 1 − co s 2 θ ( 1 − cos θ ) 2
= ( 1 − c o s θ ) 2 ( 1 − c o s θ ) ( 1 + c o s θ ) =\frac{(1-cos \theta )^{2}}{(1-cos \theta )(1+cos \theta )} = ( 1 − cos θ ) ( 1 + cos θ ) ( 1 − cos θ ) 2
( ∵ a 2 − b 2 = ( a + b ) ( a − b ) ) (∵a^{2}-b^{2}=(a+b)(a-b)) ( ∵ a 2 − b 2 = ( a + b ) ( a − b ))
= 1 − c o s θ 1 + c o s θ = R . H . S . =\frac{1-cos \theta }{1+cos \theta }=R.H.S. = 1 + cos θ 1 − cos θ = R . H . S .
∴ ∴ ∴
(ii) L.H.S. = c o s A 1 + s i n A + 1 + s i n A c o s A \frac{cosA}{1+sinA}+\frac{1+sinA}{cosA} 1 + s in A cos A + cos A 1 + s in A
= ( c o s A ) 2 + ( 1 + s i n A ) 2 ( 1 + s i n A ) ( c o s A ) =\frac{(cosA)^{2}+(1+sinA)^{2}}{(1+sinA)(cosA)} = ( 1 + s in A ) ( cos A ) ( cos A ) 2 + ( 1 + s in A ) 2
= c o s 2 A + 1 + s i n 2 A + 2 s i n A ( 1 + s i n A ) ( c o s A ) =\frac{cos^{2}A+1+sin^{2}A+2sinA}{(1+sinA)(cosA)} = ( 1 + s in A ) ( cos A ) co s 2 A + 1 + s i n 2 A + 2 s in A
( ∵ ( a + b ) 2 = a 2 + b 2 + 2 a b ) (∵(a+b)^{2}=a^{2}+b^{2}+2ab) ( ∵ ( a + b ) 2 = a 2 + b 2 + 2 ab )
= ( s i n 2 A + c o s 2 A ) + 1 + 2 s i n A ( 1 + s i n A ) ( c o s A ) =\frac{(sin^{2}A+cos^{2}A)+1+2sinA}{(1+sinA)(cosA)} = ( 1 + s in A ) ( cos A ) ( s i n 2 A + co s 2 A ) + 1 + 2 s in A
= 2 + 2 s i n A ( 1 + s i n A ) ( c o s A ) =\frac{2+2sinA}{(1+sinA)(cosA)} = ( 1 + s in A ) ( cos A ) 2 + 2 s in A
( ∵ c o s 2 A + s i n 2 A = 1 ) (∵cos^{2}A+sin^{2}A=1) ( ∵ co s 2 A + s i n 2 A = 1 )
= 2 ( 1 + s i n A ) ( 1 + s i n A ) × c o s A =\frac{2(1+sinA)}{(1+sinA) \times cosA} = ( 1 + s in A ) × cos A 2 ( 1 + s in A )
= 2 c o s A = 2 s e c A = R . H . S . =\frac{2}{cosA}=2secA=R.H.S. = cos A 2 = 2 sec A = R . H . S .
∴ L . H . S . = R . H . S . ∴L.H.S.=R.H.S. ∴ L . H . S . = R . H . S .
(iii) L . H . S . = t a n θ 1 − c o t θ + c o t θ 1 − t a n θ L.H.S.=\frac{tan \theta }{1-cot \theta }+\frac{cot \theta }{1-tan \theta } L . H . S . = 1 − co tθ t an θ + 1 − t an θ co tθ
= s i n θ c o s θ 1 − c o s θ s i n θ + c o s θ s i n θ 1 − s i n θ c o s θ =\frac{\frac{sin \theta }{cos \theta }}{1-\frac{cos \theta }{sin \theta }}+\frac{\frac{cos \theta }{sin \theta }}{1-\frac{sin \theta }{cos \theta }} = 1 − s in θ cos θ cos θ s in θ + 1 − cos θ s in θ s in θ cos θ
= s i n θ c o s θ s i n θ − c o s θ s i n θ + c o s θ s i n θ c o s θ − s i n θ c o s θ =\frac{\frac{sin \theta }{cos \theta }}{\frac{sin \theta -cos \theta }{sin \theta }}+\frac{\frac{cos \theta }{sin \theta }}{\frac{cos \theta -sin \theta }{cos \theta }} = s in θ s in θ − cos θ cos θ s in θ + cos θ cos θ − s in θ s in θ cos θ
= s i n θ × s i n θ c o s θ × ( s i n θ − c o s θ ) + c o s θ × c o s θ s i n θ × ( c o s θ − s i n θ ) =\frac{sin \theta \times sin \theta }{cos \theta \times (sin \theta -cos \theta )}+\frac{cos \theta \times cos \theta }{sin \theta \times (cos \theta -sin \theta )} = cos θ × ( s in θ − cos θ ) s in θ × s in θ + s in θ × ( cos θ − s in θ ) cos θ × cos θ
= s i n 2 θ c o s θ × ( s i n θ − c o s θ ) − c o s 2 θ s i n θ × ( s i n θ − c o s θ ) =\frac{sin^{2} \theta }{cos \theta \times (sin \theta -cos \theta )}-\frac{cos^{2} \theta }{sin \theta \times (sin \theta -cos \theta )} = cos θ × ( s in θ − cos θ ) s i n 2 θ − s in θ × ( s in θ − cos θ ) co s 2 θ
= s i n θ × s i n 2 θ − c o s θ × c o s 2 θ c o s θ × s i n θ × ( s i n θ − c o s θ ) =\frac{sin \theta \times sin^{2} \theta -cos \theta \times cos^{2} \theta }{cos \theta \times sin \theta \times (sin \theta -cos \theta )} = cos θ × s in θ × ( s in θ − cos θ ) s in θ × s i n 2 θ − cos θ × co s 2 θ
= s i n 3 θ − c o s 3 θ c o s θ × s i n θ × ( s i n θ − c o s θ ) =\frac{sin^{3} \theta -cos^{3} \theta }{cos \theta \times sin \theta \times (sin \theta -cos \theta )} = cos θ × s in θ × ( s in θ − cos θ ) s i n 3 θ − co s 3 θ
= ( s i n θ − c o s θ ) × ( s i n 2 θ + c o s 2 θ + s i n θ c o s θ ) c o s θ × s i n θ × ( s i n θ − c o s θ ) =\frac{(sin \theta -cos \theta ) \times (sin^{2} \theta +cos^{2} \theta +sin \theta cos \theta )}{cos \theta \times sin \theta \times (sin \theta -cos \theta )} = cos θ × s in θ × ( s in θ − cos θ ) ( s in θ − cos θ ) × ( s i n 2 θ + co s 2 θ + s in θ cos θ )
( ∵ a 3 − b 3 = ( a − b ) ( a 2 + b 2 + a b ) ) (∵a^{3}-b^{3}=(a-b)(a^{2}+b^{2}+ab)) ( ∵ a 3 − b 3 = ( a − b ) ( a 2 + b 2 + ab ))
= s i n 2 θ + c o s 2 θ + s i n θ c o s θ c o s θ × s i n θ =\frac{sin^{2} \theta +cos^{2} \theta +sin \theta cos \theta }{cos \theta \times sin \theta } = cos θ × s in θ s i n 2 θ + co s 2 θ + s in θ cos θ
= 1 + s i n θ c o s θ c o s θ s i n θ [ ∵ s i n 2 θ + c o s 2 θ = 1 ] =\frac{1+sin \theta cos \theta }{cos \theta sin \theta }[∵sin^{2} \theta +cos^{2} \theta =1] = cos θ s in θ 1 + s in θ cos θ [ ∵ s i n 2 θ + co s 2 θ = 1 ]
= 1 c o s θ s i n θ + 1 =\frac{1}{cos \theta sin \theta }+1 = cos θ s in θ 1 + 1
= 1 + 1 c o s θ 1 s i n θ =1+\frac{1}{cos \theta }\frac{1}{sin \theta } = 1 + cos θ 1 s in θ 1
= 1 + s e c θ c o s e c θ = R . H . S . =1+sec \theta cosec \theta =R.H.S. = 1 + sec θ cosec θ = R . H . S .
∴ L . H . S . = R . H . S . ∴L.H.S.=R.H.S. ∴ L . H . S . = R . H . S .
(iv) L . H . S . = 1 + s e c A s e c A = 1 + 1 c o s A 1 c o s A L.H.S.=\frac{1+secA}{secA}=\frac{1+\frac{1}{cosA}}{\frac{1}{cosA}} L . H . S . = sec A 1 + sec A = cos A 1 1 + cos A 1
= 1 + c o s A =1+cosA = 1 + cos A
R . H . S . = s i n 2 A 1 − c o s A ( ∵ 1 − c o s 2 A = s i n 2 A ) R.H.S.=\frac{sin^{2}A}{1-cosA}(∵1-cos^{2}A=sin^{2}A) R . H . S . = 1 − cos A s i n 2 A ( ∵ 1 − co s 2 A = s i n 2 A )
= 1 − c o s 2 A 1 − c o s A =\frac{1-cos^{2}A}{1-cosA} = 1 − cos A 1 − co s 2 A
= ( 1 + c o s A ) ( 1 − c o s A ) ( 1 ) =\frac{(1+cosA)(1-cosA)}{(1)} = ( 1 ) ( 1 + cos A ) ( 1 − cos A )
= 1 + c o s A =1+cosA = 1 + cos A
∴ L . H . S . = R . H . S . ∴L.H.S.=R.H.S. ∴ L . H . S . = R . H . S .
(v) L . H . S . = c o s A − s i n A + 1 c o s A + s i n A − 1 L.H.S.=\frac{cosA-sinA+1}{cosA+sinA-1} L . H . S . = cos A + s in A − 1 cos A − s in A + 1
(अंश और हर को s i n A sinA s in A
= c o s A s i n A − s i n A s i n A + 1 s i n A c o s A s i n A + s i n A s i n A − 1 s i n A =\frac{\frac{cosA}{sinA}-\frac{sinA}{sinA}+\frac{1}{sinA}}{\frac{cosA}{sinA}+\frac{sinA}{sinA}-\frac{1}{sinA}} = s in A cos A + s in A s in A − s in A 1 s in A cos A − s in A s in A + s in A 1
= c o t A − 1 + c o s e c A c o t A + 1 − c o s e c A =\frac{cotA-1+cosecA}{cotA+1-cosecA} = co t A + 1 − cosec A co t A − 1 + cosec A
= ( c o s e c A + c o t A ) − ( c o s e c 2 A − c o t 2 A ) ( 1 + c o t A − c o s e c A ) =\frac{(cosecA+cotA)-(cosec^{2}A-cot^{2}A)}{(1+cotA-cosecA)} = ( 1 + co t A − cosec A ) ( cosec A + co t A ) − ( cose c 2 A − co t 2 A )
( ∵ c o s e c 2 A = 1 + c o t 2 A अर्थात् c o s e c 2 A − c o t 2 A = 1 ) (∵cosec^{2}A=1+cot^{2}A \text{अर्थात्} cosec^{2}A-cot^{2}A=1) ( ∵ cose c 2 A = 1 + co t 2 A अर्थात् cose c 2 A − co t 2 A = 1 )
= ( c o s e c A + c o t A ) − ( c o s e c A + c o t A ) ( c o s e c A − c o t A ) ( 1 + c o t A − c o s e c A ) =\frac{(cosecA+cotA)-(cosecA+cotA)(cosecA-cotA)}{(1+cotA-cosecA)} = ( 1 + co t A − cosec A ) ( cosec A + co t A ) − ( cosec A + co t A ) ( cosec A − co t A )
( ∵ ( a + b ) ( a − b ) = a 2 − b 2 ) (∵(a+b)(a-b)=a^{2}-b^{2}) ( ∵ ( a + b ) ( a − b ) = a 2 − b 2 )
= ( c o s e c A + c o t A ) × 1 − ( c o s e c A − c o t A ) ( 1 + c o t A − c o s e c A ) =\frac{(cosecA+cotA) \times {1-(cosecA-cotA)}}{(1+cotA-cosecA)} = ( 1 + co t A − cosec A ) ( cosec A + co t A ) × 1 − ( cosec A − co t A )
= ( c o s e c A + c o t A ) × 1 + c o t A − c o s e c A ( 1 + c o t A − c o s e c A ) =\frac{(cosecA+cotA) \times {1+cotA-cosecA}}{(1+cotA-cosecA)} = ( 1 + co t A − cosec A ) ( cosec A + co t A ) × 1 + co t A − cosec A
= c o s e c A + c o t A =cosecA+cotA = cosec A + co t A
= R . H . S . =R.H.S. = R . H . S .
∴ L . H . S . = R . H . S . ∴L.H.S.=R.H.S. ∴ L . H . S . = R . H . S .
(vi) L . H . S . = 1 + s i n A 1 − s i n A L.H.S.=\sqrt{\frac{1+sinA}{1-sinA}} L . H . S . = 1 − s in A 1 + s in A
वर्गमूल चिह्न को हटाने के लिये अंश व हर में 1 + s i n A 1+sinA 1 + s in A
= ( 1 + s i n A ) ( 1 + s i n A ) ( 1 − s i n A ) ( 1 + s i n A ) =\sqrt{\frac{(1+sinA)(1+sinA)}{(1-sinA)(1+sinA)}} = ( 1 − s in A ) ( 1 + s in A ) ( 1 + s in A ) ( 1 + s in A )
= ( 1 + s i n A ) 2 ( 1 ) 2 − ( s i n A ) 2 =\sqrt{\frac{(1+sinA)^{2}}{(1)^{2}-(sinA)^{2}}} = ( 1 ) 2 − ( s in A ) 2 ( 1 + s in A ) 2
= ( 1 + s i n A ) 2 1 − s i n 2 A = ( 1 + s i n A ) 2 c o s 2 A =\sqrt{\frac{(1+sinA)^{2}}{1-sin^{2}A}}=\sqrt{\frac{(1+sinA)^{2}}{cos^{2}A}} = 1 − s i n 2 A ( 1 + s in A ) 2 = co s 2 A ( 1 + s in A ) 2
= 1 + s i n A c o s A = 1 c o s A + s i n A c o s A =\frac{1+sinA}{cosA}=\frac{1}{cosA}+\frac{sinA}{cosA} = cos A 1 + s in A = cos A 1 + cos A s in A
= s e c A + t a n A = R . H . S . =secA+tanA=R.H.S. = sec A + t an A = R . H . S .
∴ L . H . S . = R . H . S . ∴L.H.S.=R.H.S. ∴ L . H . S . = R . H . S .
(vii) L . H . S . = s i n θ − 2 s i n 3 θ 2 c o s 3 θ − c o s θ L.H.S.=\frac{sin \theta -2sin^{3} \theta }{2cos^{3} \theta -cos \theta } L . H . S . = 2 co s 3 θ − cos θ s in θ − 2 s i n 3 θ
= s i n θ × 1 − 2 s i n 2 θ c o s θ × 2 c o s 2 θ − 1 =\frac{sin \theta \times {1-2sin^{2} \theta }}{cos \theta \times {2cos^{2} \theta -1}} = cos θ × 2 co s 2 θ − 1 s in θ × 1 − 2 s i n 2 θ
( ∵ c o s 2 θ + s i n 2 θ = 1 ) (∵cos^{2} \theta +sin^{2} \theta =1) ( ∵ co s 2 θ + s i n 2 θ = 1 )
= s i n θ × c o s 2 θ + s i n 2 θ − 2 s i n 2 θ c o s θ × 2 c o s 2 θ − ( c o s 2 θ + s i n 2 θ ) =\frac{sin \theta \times {cos^{2} \theta +sin^{2} \theta -2sin^{2} \theta }}{cos \theta \times {2cos^{2} \theta -(cos^{2} \theta +sin^{2} \theta )}} = cos θ × 2 co s 2 θ − ( co s 2 θ + s i n 2 θ ) s in θ × co s 2 θ + s i n 2 θ − 2 s i n 2 θ
= s i n θ × c o s 2 θ − s i n 2 θ c o s θ × c o s 2 θ − s i n 2 θ =\frac{sin \theta \times {cos^{2} \theta -sin^{2} \theta }}{cos \theta \times {cos^{2} \theta -sin^{2} \theta }} = cos θ × co s 2 θ − s i n 2 θ s in θ × co s 2 θ − s i n 2 θ
= s i n θ c o s θ = t a n θ = R . H . S . =\frac{sin \theta }{cos \theta }=tan \theta =R.H.S. = cos θ s in θ = t an θ = R . H . S .
∴ L . H . S . = R . H . S . ∴L.H.S.=R.H.S. ∴ L . H . S . = R . H . S .
(viii) L . H . S . = ( s i n A + c o s e c A ) 2 + ( c o s A + s e c A ) 2 L.H.S.=(sinA+cosec A)^{2}+(cosA+secA)^{2} L . H . S . = ( s in A + cosec A ) 2 + ( cos A + sec A ) 2
= s i n 2 A + c o s e c 2 A + 2 s i n A × c o s e c A + c o s 2 A + s e c 2 A + 2 c o s A × s e c A ={sin^{2}A+cosec^{2}A+2sinA \times cosec A}+{cos^{2}A+sec^{2}A+2cosA \times secA} = s i n 2 A + cose c 2 A + 2 s in A × cosec A + co s 2 A + se c 2 A + 2 cos A × sec A
∵ c o s e c A = 1 s i n A , s e c A = 1 c o s A ∵cosec A=\frac{1}{sinA},secA=\frac{1}{cosA} ∵ cosec A = s in A 1 , sec A = cos A 1
= s i n 2 A + c o s e c 2 A + 2 s i n A × 1 s i n A + c o s 2 A + s e c 2 A + 2 c o s A × 1 c o s A =sin^{2}A+cosec^{2}A+2sinA \times \frac{1}{sinA}+cos^{2}A+sec^{2}A+2cosA \times \frac{1}{cosA} = s i n 2 A + cose c 2 A + 2 s in A × s in A 1 + co s 2 A + se c 2 A + 2 cos A × cos A 1
= s i n 2 A + c o s e c 2 A + 2 + c o s 2 A + s e c 2 A + 2 ={sin^{2}A+cosec^{2}A+2}+{cos^{2}A+sec^{2}A+2} = s i n 2 A + cose c 2 A + 2 + co s 2 A + se c 2 A + 2
= 2 + 2 + ( s i n 2 A + c o s 2 A ) + s e c 2 A + c o s e c 2 A =2+2+(sin^{2}A+cos^{2}A)+sec^{2}A+cosec^{2}A = 2 + 2 + ( s i n 2 A + co s 2 A ) + se c 2 A + cose c 2 A
= 2 + 2 + 1 + 1 + t a n 2 A + 1 + c o t 2 A =2+2+1+1+tan^{2}A+1+cot^{2}A = 2 + 2 + 1 + 1 + t a n 2 A + 1 + co t 2 A
∵ s e c 2 A = t a n 2 A + 1 , c o s e c 2 A = c o t 2 A + 1 ∵sec^{2}A=tan^{2}A+1,cosec^{2}A=cot^{2}A+1 ∵ se c 2 A = t a n 2 A + 1 , cose c 2 A = co t 2 A + 1
= 7 + t a n 2 A + c o t 2 A = R . H . S . =7+tan^{2}A+cot^{2}A=R.H.S. = 7 + t a n 2 A + co t 2 A = R . H . S .
∴ L . H . S . = R . H . S . ∴L.H.S.=R.H.S. ∴ L . H . S . = R . H . S .
(ix) L . H . S . = ( c o s e c A − s i n A ) ( s e c A − c o s A ) L.H.S.=(cosec A-sinA)(secA-cosA) L . H . S . = ( cosec A − s in A ) ( sec A − cos A )
= 1 s i n A − s i n A × 1 c o s A − c o s A =\frac{1}{sinA}-sinA \times \frac{1}{cosA}-cosA = s in A 1 − s in A × cos A 1 − cos A
= 1 − s i n 2 A s i n A × 1 − c o s 2 A c o s A =\frac{1-sin^{2}A}{sinA} \times \frac{1-cos^{2}A}{cosA} = s in A 1 − s i n 2 A × cos A 1 − co s 2 A
= c o s 2 A s i n A × s i n 2 A c o s A = s i n A c o s A =\frac{cos^{2}A}{sinA} \times \frac{sin^{2}A}{cosA}=sinAcosA = s in A co s 2 A × cos A s i n 2 A = s in A cos A
पुन: R . H . S . = 1 t a n A + c o t A R.H.S.=\frac{1}{tanA+cotA} R . H . S . = t an A + co t A 1
= 1 s i n A c o s A + c o s A s i n A =\frac{1}{\frac{sinA}{cosA}+\frac{cosA}{sinA}} = cos A s in A + s in A cos A 1
= 1 s i n 2 A + c o s 2 A s i n A c o s A =\frac{1}{\frac{sin^{2}A+cos^{2}A}{sinAcosA}} = s in A cos A s i n 2 A + co s 2 A 1
= s i n A c o s A s i n 2 A + c o s 2 A =\frac{sinAcosA}{sin^{2}A+cos^{2}A} = s i n 2 A + co s 2 A s in A cos A
= s i n A c o s A 1 = s i n A c o s A =\frac{sinAcosA}{1}=sinAcosA = 1 s in A cos A = s in A cos A
∴ L . H . S . = R . H . S . ∴L.H.S.=R.H.S. ∴ L . H . S . = R . H . S .
(x) L . H . S . = 1 + t a n 2 A 1 + c o t 2 A = s e c 2 A c o s e c 2 A L.H.S.=\frac{1+tan^{2}A}{1+cot^{2}A}=\frac{sec^{2}A}{cosec^{2}A} L . H . S . = 1 + co t 2 A 1 + t a n 2 A = cose c 2 A se c 2 A
∵ 1 + t a n 2 A = s e c 2 A और 1 + c o t 2 A = c o s e c 2 A ∵1+tan^{2}A=sec^{2}A \text{और} 1+cot^{2}A=cosec^{2}A ∵ 1 + t a n 2 A = se c 2 A और 1 + co t 2 A = cose c 2 A
= 1 c o s 2 A 1 s i n 2 A = s i n 2 A c o s 2 A = t a n 2 A =\frac{\frac{1}{cos^{2}A}}{\frac{1}{sin^{2}A}}=\frac{sin^{2}A}{cos^{2}A}=tan^{2}A = s i n 2 A 1 co s 2 A 1 = co s 2 A s i n 2 A = t a n 2 A
R . H . S . = 1 − t a n A 1 − c o t A 2 = 1 − s i n A c o s A 1 − c o s A s i n A 2 R.H.S.=\frac{1-tanA}{1-cotA}^{2}={\frac{1-\frac{sinA}{cosA}}{1-\frac{cosA}{sinA}}}^{2} R . H . S . = 1 − co t A 1 − t an A 2 = 1 − s in A cos A 1 − cos A s in A 2
= c o s A − s i n A c o s A s i n A − c o s A s i n A 2 ={\frac{\frac{cosA-sinA}{cosA}}{\frac{sinA-cosA}{sinA}}}^{2} = s in A s in A − cos A cos A cos A − s in A 2
= − ( s i n A − c o s A ) c o s A × s i n A ( s i n A ) 2 ={\frac{-(sinA-cosA)}{cosA} \times \frac{sinA}{(sinA)}}^{2} = cos A − ( s in A − cos A ) × ( s in A ) s in A 2
= ( − s i n A ) 2 c o s 2 A = s i n 2 A c o s 2 A = t a n 2 A =\frac{(-sinA)^{2}}{cos^{2}A}=\frac{sin^{2}A}{cos^{2}A}=tan^{2}A = co s 2 A ( − s in A ) 2 = co s 2 A s i n 2 A = t a n 2 A
∴ ∴ ∴
भाग 3 — परीक्षा उपयोगी प्रश्न (Important Q&A)
वस्तुनिष्ठ प्रश्न (MCQs), अतिलघूत्तरात्मक, लघूत्तरात्मक एवं निबंधात्मक परीक्षा उपयोगी प्रश्नोत्तर।
--- अन्य महत्त्वपूर्ण प्रश्न ---
बहुविकल्पीय प्रश्न-
1. c o s e c 2 45 ∘ − c o t 2 45 ∘ cosec^{2}45^{∘}-cot^{2}45^{∘} cose c 2 4 5 ∘ − co t 2 4 5 ∘
2. यदि c o s A = 12 13 cosA=\frac{12}{13} cos A = 13 12 s i n A sinA s in A
(A) 13 12 \frac{13}{12} 12 13
3. यदि s i n A = 1 2 sinA=\frac{1}{2} s in A = 2 1 2 s i n A c o s A 2sinAcosA 2 s in A cos A
(B) 3 2 \frac{\sqrt{3}}{2} 2 3 (D) 1 2 \frac{1}{\sqrt{2}} 2 1
4. 2 t a n 30 ∘ 1 − t a n 2 30 ∘ \frac{2tan30^{∘}}{1-tan^{2}30^{∘}} 1 − t a n 2 3 0 ∘ 2 t an 3 0 ∘
(A) 1 3 \frac{1}{\sqrt{3}} 3 1
5. t a n 45 ∘ + c o t 45 ∘ tan45^{∘}+cot45^{∘} t an 4 5 ∘ + co t 4 5 ∘
6. c o s ( 90 ∘ − 48 ∘ ) cos(90^{∘}-48^{∘}) cos ( 9 0 ∘ − 4 8 ∘ )
(A) s e c 48 ∘ sec48^{∘} sec 4 8 ∘ (B) t a n 48 ∘ tan48^{∘} t an 4 8 ∘ (C) s i n 48 ∘ sin48^{∘} s in 4 8 ∘ (D) c o t 48 ∘ cot48^{∘} co t 4 8 ∘
7. t a n θ 1 + t a n 2 θ \frac{tan \theta }{\sqrt{1+tan^{2} \theta }} 1 + t a n 2 θ t an θ
8. c o s e c 2 θ − 1 c o s e c θ \frac{\sqrt{cosec^{2} \theta -1}}{cosec \theta } cosec θ cose c 2 θ − 1
(D) c o s e c θ cosec \theta cosec θ
9. s i n θ c o s e c θ + c o s θ s e c θ sin \theta cosec \theta +cos \theta sec \theta s in θ cosec θ + cos θ sec θ
10. 2 c o s e c 30 ∘ s e c 30 ∘ 2cosec30^{∘}sec30^{∘} 2 cosec 3 0 ∘ sec 3 0 ∘
(A) 2 3 \frac{2}{\sqrt{3}} 3 2 (B) 3 2 \frac{\sqrt{3}}{2} 2 3 (D) 8 3 \frac{8}{\sqrt{3}} 3 8
11. यदि s i n θ = 3 2 sin \theta =\frac{\sqrt{3}}{2} s in θ = 2 3 θ \theta θ
12. 2 s i n 45 ∘ c o s 45 ∘ 2sin45^{∘}cos45^{∘} 2 s in 4 5 ∘ cos 4 5 ∘
(A) 1 2 \frac{1}{\sqrt{2}} 2 1
13. यदि c o s ( 90 ∘ − θ ) = 1 2 cos(90^{∘}- \theta )=\frac{1}{2} cos ( 9 0 ∘ − θ ) = 2 1 θ \theta θ
14. s i n 2 50 ∘ + c o s 2 50 ∘ + 1 sin^{2}50^{∘}+cos^{2}50^{∘}+1 s i n 2 5 0 ∘ + co s 2 5 0 ∘ + 1
15. 1 − t a n 2 45 ∘ 1 + t a n 2 45 ∘ \frac{1-tan^{2}45^{∘}}{1+tan^{2}45^{∘}} 1 + t a n 2 4 5 ∘ 1 − t a n 2 4 5 ∘
(A) t a n 90 ∘ tan90^{∘} t an 9 0 ∘ (C) s i n 45 ∘ sin45^{∘} s in 4 5 ∘
16. 2 s i n 2 60 ∘ c o s 60 ∘ 2sin^{2}60^{∘}cos60^{∘} 2 s i n 2 6 0 ∘ cos 6 0 ∘
17. यदि 8 t a n x = 15 8tanx=15 8 t an x = 15 s i n x − c o s x sinx-cosx s in x − cos x
18. यदि t a n θ = a b tan \theta =\frac{a}{b} t an θ = b a a s i n θ + b c o s θ a s i n θ − b c o s θ \frac{asin \theta +bcos \theta }{asin \theta -bcos \theta } a s in θ − b cos θ a s in θ + b cos θ
(A) a 2 + b 2 a 2 − b 2 \frac{a^{2}+b^{2}}{a^{2}-b^{2}} a 2 − b 2 a 2 + b 2 (B) a 2 − b 2 a 2 + b 2 \frac{a^{2}-b^{2}}{a^{2}+b^{2}} a 2 + b 2 a 2 − b 2 (C) a + b a − b \frac{a+b}{a-b} a − b a + b (D) a − b a + b \frac{a-b}{a+b} a + b a − b
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19. यदि θ = 45 ∘ \theta =45^{∘} θ = 4 5 ∘ 1 − c o s 2 θ s i n 2 θ \frac{1-cos2 \theta }{sin2 \theta } s in 2 θ 1 − cos 2 θ
( माध्य. शिक्षा बोर्ड, मॉडल पेपर, 2024-25 )
(A) शून्य
(B) 1
(C) 2
(D) अनन्त
20. यदि c s c θ = 2 3 csc \theta =\frac{2}{\sqrt{3}} csc θ = 3 2 θ \theta θ
21. s i n θ 1 + c o s θ \frac{sin \theta }{1+cos \theta } 1 + cos θ s in θ
(A) 1 + c o s θ s i n θ \frac{1+cos \theta }{sin \theta } s in θ 1 + cos θ (B) 1 − c o s θ c o s θ \frac{1-cos \theta }{cos \theta } cos θ 1 − cos θ (C) 1 − c o s θ s i n θ \frac{1-cos \theta }{sin \theta } s in θ 1 − cos θ (D) 1 − s i n θ c o s θ \frac{1-sin \theta }{cos \theta } cos θ 1 − s in θ
22. यदि s e c θ + t a n θ = x sec \theta +tan \theta =x sec θ + t an θ = x s e c θ = sec \theta = sec θ =
(A) x 2 + 1 x \frac{x^{2}+1}{x} x x 2 + 1 (B) x 2 + 1 2 x \frac{x^{2}+1}{2x} 2 x x 2 + 1 (C) x 2 − 1 2 x \frac{x^{2}-1}{2x} 2 x x 2 − 1 (D) x 2 − 1 x \frac{x^{2}-1}{x} x x 2 − 1
23. 8 s e c 2 A − 8 t a n 2 A 8sec^{2}A-8tan^{2}A 8 se c 2 A − 8 t a n 2 A
24. 1 + c o t 2 θ 1 + t a n 2 θ \frac{1+cot^{2} \theta }{1+tan^{2} \theta } 1 + t a n 2 θ 1 + co t 2 θ
(A) c o t 2 θ cot^{2} \theta co t 2 θ (C) s e c 2 θ sec^{2} \theta se c 2 θ (D) t a n 2 θ tan^{2} \theta t a n 2 θ
25. 1 + s i n θ 1 − s i n θ \sqrt{\frac{1+sin \theta }{1-sin \theta }} 1 − s in θ 1 + s in θ
(A) s e c θ + t a n θ sec \theta +tan \theta sec θ + t an θ (B) s e c θ − t a n θ sec \theta -tan \theta sec θ − t an θ (C) s e c 2 θ + t a n 2 θ sec^{2} \theta +tan^{2} \theta se c 2 θ + t a n 2 θ (D) s e c 2 θ − t a n 2 θ sec^{2} \theta -tan^{2} \theta se c 2 θ − t a n 2 θ
26. t a n 2 60 ∘ tan^{2}60^{∘} t a n 2 6 0 ∘
27. c o s 2 45 ∘ cos^{2}45^{∘} co s 2 4 5 ∘
(A) 1 2 \frac{1}{\sqrt{2}} 2 1 (B) 3 2 \frac{\sqrt{3}}{2} 2 3 (D) 1 3 \frac{1}{\sqrt{3}} 3 1
28. 3 s e c 45 ∘ c o s 45 ∘ 3sec45^{∘}cos45^{∘} 3 sec 4 5 ∘ cos 4 5 ∘
( माध्य. शिक्षा बोर्ड, 2023 )
29. Δ A B C \Delta ABC Δ A B C ∠ B \angle B ∠ B c o s A = 3 5 cosA=\frac{3}{5} cos A = 5 3 s i n A sinA s in A
( माध्य. शिक्षा बोर्ड, मॉडल पेपर, 2023-24 )
30. 2 s i n 45 ∘ c o s 45 ∘ 2sin45^{∘}cos45^{∘} 2 s in 4 5 ∘ cos 4 5 ∘
( माध्य. शिक्षा बोर्ड, 2024 )
31. 2 s i n θ c s c θ 2sin \theta csc \theta 2 s in θ csc θ
32. s i n 45 ∘ c o s 45 ∘ \frac{sin45^{∘}}{cos45^{∘}} cos 4 5 ∘ s in 4 5 ∘
33. यदि t a n θ = 5 12 tan \theta =\frac{5}{12} t an θ = 12 5 s e c θ sec \theta sec θ
( प्रश्न बैंक )
(A) 12 13 \frac{12}{13} 13 12 (D) 13 12 \frac{13}{12} 12 13
34. यदि s e c θ = 41 40 sec \theta =\frac{41}{40} sec θ = 40 41 c o t θ + 1 cot \theta +1 co tθ + 1
( प्रश्न बैंक )
35. यदि s i n A = 3 4 sinA=\frac{3}{4} s in A = 4 3 c s c A cscA csc A
( प्रश्न बैंक )
36. यदि t a n 3 x = 1 tan3x=1 t an 3 x = 1 x x x
( प्रश्न बैंक )
37. Δ A B C \Delta ABC Δ A B C ∠ B \angle B ∠ B c o s θ = 3 5 cos \theta =\frac{3}{5} cos θ = 5 3 s i n A sinA s in A
उत्तरमाला
1. (B)
2. (C)
3. (B)
4. (D)
5. (B)
6. (C)
7. (B)
8. (A)
9. (A)
10. (D)
11. (C)
12. (C)
13. (D)
14. (A)
15. (D)
16. (C)
17. (D)
18. (A)
19. (B)
20. (B)
21. (C)
22. (B)
23. (C)
24. (A)
25. (A)
26. (A)
27. (C)
28. (D)
29. (B)
30. (D)
31. (C)
32. (B)
33. (D)
34. (C)
35. (C)
36. (A)
37. (B)
रिक्त स्थानों की पूर्ति कीजिए-
1. 2 s i n 2 60 ∘ c o s 60 ∘ 2sin^{2}60^{∘}cos60^{∘} 2 s i n 2 6 0 ∘ cos 6 0 ∘
2. t a n 30 ∘ t a n 60 ∘ tan30^{∘}tan60^{∘} t an 3 0 ∘ t an 6 0 ∘
3. s i n 2 θ + 1 1 + t a n 2 θ sin^{2} \theta +\frac{1}{1+tan^{2} \theta } s i n 2 θ + 1 + t a n 2 θ 1
4. ( 1 + t a n 2 θ ) ( 1 − s i n θ ) ( 1 + s i n θ ) (1+tan^{2} \theta )(1-sin \theta )(1+sin \theta ) ( 1 + t a n 2 θ ) ( 1 − s in θ ) ( 1 + s in θ )
5. 9 s e c 2 θ − 9 t a n 2 θ = 9sec^{2} \theta -9tan^{2} \theta = 9 se c 2 θ − 9 t a n 2 θ =
6. यदि s i n θ = c o s θ , 0 ≤ θ ≤ 90 ∘ sin \theta =cos \theta ,0 \le \theta \le 90^{∘} s in θ = cos θ , 0 ≤ θ ≤ 9 0 ∘ θ \theta θ
7. 5 c o t 2 θ − 5 c s c 2 θ 5cot^{2} \theta -5csc^{2} \theta 5 co t 2 θ − 5 cs c 2 θ
8. यदि s i n θ − c o s θ = 0 sin \theta -cos \theta =0 s in θ − cos θ = 0 s i n 4 θ + c o s 4 θ sin^{4} \theta +cos^{4} \theta s i n 4 θ + co s 4 θ
9. s i n 2 3 θ + c o s 2 3 θ sin^{2}3 \theta +cos^{2}3 \theta s i n 2 3 θ + co s 2 3 θ
10. यदि c o s θ = 3 2 cos \theta =\frac{\sqrt{3}}{2} cos θ = 2 3 θ \theta θ
11. यदि s i n A + s i n 2 A = 1 sinA+sin^{2}A=1 s in A + s i n 2 A = 1 ( c o s 2 A ) (cos^{2}A) ( co s 2 A )
12. c o s 2 45 ∘ cos^{2}45^{∘} co s 2 4 5 ∘
13. c o s 2 60 ∘ + s i n 2 60 ∘ cos^{2}60^{∘}+sin^{2}60^{∘} co s 2 6 0 ∘ + s i n 2 6 0 ∘
14. c o s 2 60 ∘ − s i n 2 60 ∘ cos^{2}60^{∘}-sin^{2}60^{∘} co s 2 6 0 ∘ − s i n 2 6 0 ∘
15. s i n 2 θ + c o s 2 θ sin^{2} \theta +cos^{2} \theta s i n 2 θ + co s 2 θ
16. 1 − c o s 2 θ \sqrt{1-cos^{2} \theta } 1 − co s 2 θ
17. t a n 2 60 ∘ tan^{2}60^{∘} t a n 2 6 0 ∘
18. t a n θ = 3 4 tan \theta =\frac{3}{4} t an θ = 4 3 s i n θ sin \theta s in θ
19. 2 t a n 30 ∘ 1 + t a n 2 30 ∘ \frac{2tan30^{∘}}{1+tan^{2}30^{∘}} 1 + t a n 2 3 0 ∘ 2 t an 3 0 ∘
20. 2 t a n 30 ∘ 1 − t a n 2 30 ∘ \frac{2tan30^{∘}}{1-tan^{2}30^{∘}} 1 − t a n 2 3 0 ∘ 2 t an 3 0 ∘
उत्तर— 1. 3 4 \frac{3}{4} 4 3 2. 1 3. 1 4. 1 5. 9 6. 45 ∘ 45^{∘} 4 5 ∘ 7. − 5 -5 − 5 8. 1 2 \frac{1}{2} 2 1 9. 1 10. 30 ∘ 30^{∘} 3 0 ∘ 11. 1 12. 1 2 \frac{1}{2} 2 1 13. 1 14. − 1 2 -\frac{1}{2} − 2 1 15. 1 16. s i n θ sin \theta s in θ 17. 3 18. 3 5 \frac{3}{5} 5 3 19. 3 2 \frac{\sqrt{3}}{2} 2 3 20. 3 \sqrt{3} 3
अतिलघूत्तरात्मक प्रश्न—
प्रश्न 1. 2 t a n 2 45 ∘ + c o s 2 30 ∘ − s i n 2 60 ∘ 2tan^{2}45^{∘}+cos^{2}30^{∘}-sin^{2}60^{∘} 2 t a n 2 4 5 ∘ + co s 2 3 0 ∘ − s i n 2 6 0 ∘ का मान ज्ञात कीजिए। (माध्य. शिक्षा बोर्ड, 2022)
हल— 2 t a n 2 45 ∘ + c o s 2 30 ∘ − s i n 2 60 ∘ 2tan^{2}45^{∘}+cos^{2}30^{∘}-sin^{2}60^{∘} 2 t a n 2 4 5 ∘ + co s 2 3 0 ∘ − s i n 2 6 0 ∘ मान रखने पर-
⇒ 2 × ( 1 ) 2 + 3 2 2 − 3 2 2 ⇒2 \times (1)^{2}+\frac{\sqrt{3}}{2}^{2}-\frac{\sqrt{3}}{2}^{2} ⇒ 2 × ( 1 ) 2 + 2 3 2 − 2 3 2
⇒ 2 × 1 + 3 4 − 3 4 = 2 + 0 = 2 ⇒2 \times 1+\frac{3}{4}-\frac{3}{4}=2+0=2 ⇒ 2 × 1 + 4 3 − 4 3 = 2 + 0 = 2
प्रश्न 2. यदि s i n 3 x = 1 sin3x=1 s in 3 x = 1 तो x x x का मान ज्ञात कीजिए। (माध्य. शिक्षा बोर्ड, मॉडल पेपर, 2021-22)
हल— s i n 3 x = 1 = s i n 90 ∘ sin3x=1=sin90^{∘} s in 3 x = 1 = s in 9 0 ∘
⇒ 3 x = 90 ∘ ⇒3x=90^{∘} ⇒ 3 x = 9 0 ∘
⇒ x = 30 ∘ ⇒x=30^{∘} ⇒ x = 3 0 ∘
प्रश्न 3. मान ज्ञात कीजिए— c o s 30 ∘ − s i n 60 ∘ cos30^{∘}-sin60^{∘} cos 3 0 ∘ − s in 6 0 ∘ (माध्य. शिक्षा बोर्ड, मॉडल पेपर, 2021-22)
हल— c o s 30 ∘ − s i n 60 ∘ = 3 2 − 3 2 = 0 cos30^{∘}-sin60^{∘}=\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}=0 cos 3 0 ∘ − s in 6 0 ∘ = 2 3 − 2 3 = 0
प्रश्न 4. s i n 2 60 ∘ + c o s 2 30 ∘ sin^{2}60^{∘}+cos^{2}30^{∘} s i n 2 6 0 ∘ + co s 2 3 0 ∘ का मान लिखिए।
हल— 3 2 2 + 3 2 2 = 3 4 + 3 4 \frac{\sqrt{3}}{2}^{2}+\frac{\sqrt{3}}{2}^{2}=\frac{3}{4}+\frac{3}{4} 2 3 2 + 2 3 2 = 4 3 + 4 3
= 6 4 = 3 2 =\frac{6}{4}=\frac{3}{2} = 4 6 = 2 3
प्रश्न 5. 1 − s i n 2 40 ∘ c o s 40 ∘ \frac{\sqrt{1-sin^{2}40^{∘}}}{cos40^{∘}} cos 4 0 ∘ 1 − s i n 2 4 0 ∘ का सरलतम मान लिखिए।
हल— c o s 2 40 ∘ c o s 40 ∘ = c o s 40 ∘ c o s 40 ∘ = 1 \frac{\sqrt{cos^{2}40^{∘}}}{cos40^{∘}}=\frac{cos40^{∘}}{cos40^{∘}}=1 cos 4 0 ∘ co s 2 4 0 ∘ = cos 4 0 ∘ cos 4 0 ∘ = 1 उत्तर
प्रश्न 6. 3 s i n 60 ∘ − 4 s i n 3 60 ∘ 3sin60^{∘}-4sin^{3}60^{∘} 3 s in 6 0 ∘ − 4 s i n 3 6 0 ∘ का मान ज्ञात कीजिए।
हल— 3 s i n 60 ∘ − 4 s i n 3 60 ∘ 3sin60^{∘}-4sin^{3}60^{∘} 3 s in 6 0 ∘ − 4 s i n 3 6 0 ∘
⇒ 3 × 3 2 − 4 3 2 3 ⇒3 \times \frac{\sqrt{3}}{2}-4\frac{\sqrt{3}}{2}^{3} ⇒ 3 × 2 3 − 4 2 3 3
⇒ 3 3 2 − 4 × 3 3 8 ⇒\frac{3\sqrt{3}}{2}-\frac{4 \times 3\sqrt{3}}{8} ⇒ 2 3 3 − 8 4 × 3 3
⇒ 3 3 2 − 3 3 2 = 0 ⇒\frac{3\sqrt{3}}{2}-\frac{3\sqrt{3}}{2}=0 ⇒ 2 3 3 − 2 3 3 = 0
प्रश्न 7. s i n θ ⋅ c o s e c θ − c o s θ ⋅ s e c θ sin \theta ·cosec \theta -cos \theta ·sec \theta s in θ ⋅ cosec θ − cos θ ⋅ sec θ का मान ज्ञात कीजिए।
हल— s i n θ ⋅ c o s e c θ − c o s θ ⋅ s e c θ sin \theta ·cosec \theta -cos \theta ·sec \theta s in θ ⋅ cosec θ − cos θ ⋅ sec θ
= s i n θ ⋅ 1 s i n θ − c o s θ ⋅ 1 c o s θ = 1 − 1 = 0 =sin \theta ·\frac{1}{sin \theta }-cos \theta ·\frac{1}{cos \theta }=1-1=0 = s in θ ⋅ s in θ 1 − cos θ ⋅ cos θ 1 = 1 − 1 = 0
प्रश्न 8. ( 1 − s i n 2 θ ) s e c 2 θ (1-sin^{2} \theta )sec^{2} \theta ( 1 − s i n 2 θ ) se c 2 θ का मान लिखिए।
हल— ( 1 − s i n 2 θ ) s e c 2 θ (1-sin^{2} \theta )sec^{2} \theta ( 1 − s i n 2 θ ) se c 2 θ
= c o s 2 θ ⋅ s e c 2 θ =cos^{2} \theta ·sec^{2} \theta = co s 2 θ ⋅ se c 2 θ
= c o s 2 θ ⋅ 1 c o s 2 θ = 1 =cos^{2} \theta ·\frac{1}{cos^{2} \theta }=1 = co s 2 θ ⋅ co s 2 θ 1 = 1
प्रश्न 9. 1 c o s e c 2 θ − 1 \frac{1}{\sqrt{cosec^{2} \theta -1}} cose c 2 θ − 1 1 का मान लिखिए।
हल— 1 c o s e c 2 θ − 1 = 1 c o t 2 θ = 1 c o t θ \frac{1}{\sqrt{cosec^{2} \theta -1}}=\frac{1}{\sqrt{cot^{2} \theta }}=\frac{1}{cot \theta } cose c 2 θ − 1 1 = co t 2 θ 1 = co tθ 1
= s i n θ c o s θ = t a n θ =\frac{sin \theta }{cos \theta }=tan \theta = cos θ s in θ = t an θ
प्रश्न 10. c o s θ s i n θ × c o t θ × t a n 2 θ \frac{cos \theta }{sin \theta } \times cot \theta \times tan^{2} \theta s in θ cos θ × co tθ × t a n 2 θ का मान लिखिए।
हल— c o s θ s i n θ × c o s θ s i n θ × s i n 2 θ c o s 2 θ \frac{cos \theta }{sin \theta } \times \frac{cos \theta }{sin \theta } \times \frac{sin^{2} \theta }{cos^{2} \theta } s in θ cos θ × s in θ cos θ × co s 2 θ s i n 2 θ
= c o s 2 θ s i n 2 θ × s i n 2 θ c o s 2 θ = 1 =\frac{cos^{2} \theta }{sin^{2} \theta } \times \frac{sin^{2} \theta }{cos^{2} \theta }=1 = s i n 2 θ co s 2 θ × co s 2 θ s i n 2 θ = 1
प्रश्न 11. 2 s i n 2 A ( 1 + c o t 2 A ) 2sin^{2}A(1+cot^{2}A) 2 s i n 2 A ( 1 + co t 2 A ) का मान लिखिए।
हल— 2 s i n 2 A × c o s e c 2 A 2sin^{2}A \times cosec^{2}A 2 s i n 2 A × cose c 2 A
= 2 s i n 2 A × 1 s i n 2 A = 2 =2sin^{2}A \times \frac{1}{sin^{2}A}=2 = 2 s i n 2 A × s i n 2 A 1 = 2
प्रश्न 12. त्रिकोणमितीय अनुपात t a n A tanA t an A को s e c A secA sec A के पदों में लिखिए।
हल— 1 + t a n 2 A = s e c 2 A 1+tan^{2}A=sec^{2}A 1 + t a n 2 A = se c 2 A
t a n 2 A = s e c 2 A − 1 tan^{2}A=sec^{2}A-1 t a n 2 A = se c 2 A − 1
( t a n A ) 2 = s e c 2 A − 1 ) (tanA)^{2}=sec^{2}A-1) ( t an A ) 2 = se c 2 A − 1 )
∴ t a n A = ± s e c 2 A − 1 ∴tanA= \pm \sqrt{sec^{2}A-1} ∴ t an A = ± se c 2 A − 1
[न्यून कोण A A A
अर्थात् t a n A = s e c 2 A − 1 tanA=\sqrt{sec^{2}A-1} t an A = se c 2 A − 1
प्रश्न 13. t a n 2 60 ∘ + 3 c o s 2 30 ∘ tan^{2}60^{∘}+3cos^{2}30^{∘} t a n 2 6 0 ∘ + 3 co s 2 3 0 ∘ का मान ज्ञात कीजिए। (प्रश्न बैंक)
हल— t a n 2 60 ∘ + 3 c o s 2 30 ∘ tan^{2}60^{∘}+3cos^{2}30^{∘} t a n 2 6 0 ∘ + 3 co s 2 3 0 ∘
त्रिकोणमितीय अनुपातों का मान रखने पर
( 3 ) 2 ) (\sqrt{3})^{2}) ( 3 ) 2 )
⇒ 3 + 3 × 3 4 = 12 + 9 4 = 21 4 ⇒3+\frac{3 \times 3}{4}=\frac{12+9}{4}=\frac{21}{4} ⇒ 3 + 4 3 × 3 = 4 12 + 9 = 4 21
प्रश्न 14. s i n 60 ∘ c o s 30 ∘ + s i n 30 ∘ c o s 60 ∘ sin60^{∘}cos30^{∘}+sin30^{∘}cos60^{∘} s in 6 0 ∘ cos 3 0 ∘ + s in 3 0 ∘ cos 6 0 ∘ का मान ज्ञात कीजिए।
हल— s i n 60 ∘ c o s 30 ∘ + s i n 30 ∘ c o s 60 ∘ sin60^{∘}cos30^{∘}+sin30^{∘}cos60^{∘} s in 6 0 ∘ cos 3 0 ∘ + s in 3 0 ∘ cos 6 0 ∘
⇒ 3 2 ⋅ 3 2 + 1 2 ⋅ 1 2 ⇒\frac{\sqrt{3}}{2}·\frac{\sqrt{3}}{2}+\frac{1}{2}·\frac{1}{2} ⇒ 2 3 ⋅ 2 3 + 2 1 ⋅ 2 1
⇒ 3 4 + 1 4 = 4 4 = 1 ⇒\frac{3}{4}+\frac{1}{4}=\frac{4}{4}=1 ⇒ 4 3 + 4 1 = 4 4 = 1
प्रश्न 15. c o s 45 ∘ c o s 60 ∘ − s i n 45 ∘ s i n 60 ∘ cos45^{∘}cos60^{∘}-sin45^{∘}sin60^{∘} cos 4 5 ∘ cos 6 0 ∘ − s in 4 5 ∘ s in 6 0 ∘ का मान ज्ञात कीजिए।
हल— c o s 45 ∘ c o s 60 ∘ − s i n 45 ∘ s i n 60 ∘ cos45^{∘}cos60^{∘}-sin45^{∘}sin60^{∘} cos 4 5 ∘ cos 6 0 ∘ − s in 4 5 ∘ s in 6 0 ∘
⇒ 1 2 × 1 2 − 1 2 × 3 2 ⇒\frac{1}{\sqrt{2}} \times \frac{1}{2}-\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} ⇒ 2 1 × 2 1 − 2 1 × 2 3
⇒ 1 2 2 − 3 2 2 = 1 − 3 2 2 ⇒\frac{1}{2\sqrt{2}}-\frac{\sqrt{3}}{2\sqrt{2}}=\frac{1-\sqrt{3}}{2\sqrt{2}} ⇒ 2 2 1 − 2 2 3 = 2 2 1 − 3
प्रश्न 16. 2 t a n 30 ∘ 1 − t a n 2 30 ∘ \frac{2tan30^{∘}}{1-tan^{2}30^{∘}} 1 − t a n 2 3 0 ∘ 2 t an 3 0 ∘ का मान ज्ञात कीजिए। (माध्य. शिक्षा बोर्ड, 2023)
हल— 2 t a n 30 ∘ 1 − t a n 2 30 ∘ \frac{2tan30^{∘}}{1-tan^{2}30^{∘}} 1 − t a n 2 3 0 ∘ 2 t an 3 0 ∘
⇒ 2 × 1 3 1 − 1 3 2 = 2 3 1 − 1 3 = 2 3 2 3 ⇒\frac{2 \times \frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}^{2}}=\frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}}=\frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} ⇒ 1 − 3 1 2 2 × 3 1 = 1 − 3 1 3 2 = 3 2 3 2
= 2 3 × 3 2 = 3 =\frac{2}{\sqrt{3}} \times \frac{3}{2}=\sqrt{3} = 3 2 × 2 3 = 3
प्रश्न 17. यदि c o t θ = 12 5 cot \theta =\frac{12}{5} co tθ = 5 12 है तब s i n θ sin \theta s in θ का मान लिखिए।
हल— ( A C ) 2 = ( 12 ) 2 + ( 5 ) 2 (AC)^{2}=(12)^{2}+(5)^{2} ( A C ) 2 = ( 12 ) 2 + ( 5 ) 2
A C 2 = 144 + 25 = 169 AC^{2}=144+25=169 A C 2 = 144 + 25 = 169
∴ A C = 169 = 13 ∴AC=\sqrt{169}=13 ∴ A C = 169 = 13
∴ s i n θ = 5 13 ∴sin \theta =\frac{5}{13} ∴ s in θ = 13 5
[Right angled triangle ABC with angle at A, AB=12, BC=5, AC=13]
प्रश्न 18. t a n 2 60 ∘ + s i n 2 45 ∘ tan^{2}60^{∘}+sin^{2}45^{∘} t a n 2 6 0 ∘ + s i n 2 4 5 ∘ का मान लिखिए।
हल— t a n 2 60 ∘ + s i n 2 45 ∘ tan^{2}60^{∘}+sin^{2}45^{∘} t a n 2 6 0 ∘ + s i n 2 4 5 ∘
⇒ ( 3 ) 2 + 1 2 2 ⇒(\sqrt{3})^{2}+\frac{1}{\sqrt{2}}^{2} ⇒ ( 3 ) 2 + 2 1 2
⇒ 3 + 1 2 = 7 2 ⇒3+\frac{1}{2}=\frac{7}{2} ⇒ 3 + 2 1 = 2 7
प्रश्न 19. 2 s i n 2 60 ∘ + 3 c o t 2 30 ∘ − t a n 45 ∘ 2sin^{2}60^{∘}+3cot^{2}30^{∘}-tan45^{∘} 2 s i n 2 6 0 ∘ + 3 co t 2 3 0 ∘ − t an 4 5 ∘ का मान ज्ञात कीजिए।
हल— 2 s i n 2 60 ∘ + 3 c o t 2 30 ∘ − t a n 45 ∘ 2sin^{2}60^{∘}+3cot^{2}30^{∘}-tan45^{∘} 2 s i n 2 6 0 ∘ + 3 co t 2 3 0 ∘ − t an 4 5 ∘
त्रिकोणमितीय कोण के मान रखने पर
⇒ 2 × 3 2 2 + 3 ( 3 ) 2 − 1 ⇒2 \times \frac{\sqrt{3}}{2}^{2}+3(\sqrt{3})^{2}-1 ⇒ 2 × 2 3 2 + 3 ( 3 ) 2 − 1
⇒ 2 × 3 4 + 3 × 3 − 1 ⇒2 \times \frac{3}{4}+3 \times 3-1 ⇒ 2 × 4 3 + 3 × 3 − 1
= 3 2 + 9 − 1 = 3 2 + 8 =\frac{3}{2}+9-1=\frac{3}{2}+8 = 2 3 + 9 − 1 = 2 3 + 8
= 16 + 3 2 = 19 2 =\frac{16+3}{2}=\frac{19}{2} = 2 16 + 3 = 2 19
प्रश्न 20. यदि c o s A = 3 2 cosA=\frac{\sqrt{3}}{2} cos A = 2 3 , 0 ∘ < A < 90 ∘ 0^{∘}<A<90^{∘} 0 ∘ < A < 9 0 ∘ है, तो कोण A A A का मान ज्ञात कीजिए।
हल— c o s A = 3 2 cosA=\frac{\sqrt{3}}{2} cos A = 2 3
⇒ c o s A = c o s 30 ∘ ⇒cosA=cos30^{∘} ⇒ cos A = cos 3 0 ∘
⇒ A = 30 ∘ ⇒A=30^{∘} ⇒ A = 3 0 ∘
प्रश्न 21. यदि s i n α = 1 2 sin \alpha =\frac{1}{2} s in α = 2 1 और c o s β = 1 2 cos \beta =\frac{1}{2} cos β = 2 1 तब ( α ) ( \alpha ) ( α ) का मान ज्ञात कीजिए।
हल— s i n α = 1 2 = s i n 30 ∘ sin \alpha =\frac{1}{2}=sin30^{∘} s in α = 2 1 = s in 3 0 ∘
α = 30 ∘ \alpha =30^{∘} α = 3 0 ∘
c o s β = 1 2 = c o s 60 ∘ cos \beta =\frac{1}{2}=cos60^{∘} cos β = 2 1 = cos 6 0 ∘
β = 60 ∘ \beta =60^{∘} β = 6 0 ∘
∴ α + β = 30 ∘ + 60 ∘ = 90 ∘ ∴ \alpha + \beta =30^{∘}+60^{∘}=90^{∘} ∴ α + β = 3 0 ∘ + 6 0 ∘ = 9 0 ∘
प्रश्न 22. यदि s i n θ = 1 2 sin \theta =\frac{1}{2} s in θ = 2 1 हो तो 1 − 2 s i n 2 θ s i n θ \frac{1-2sin^{2} \theta }{sin \theta } s in θ 1 − 2 s i n 2 θ का मान ज्ञात कीजिये।
हल— s i n θ = 1 2 = s i n 30 ∘ sin \theta =\frac{1}{2}=sin30^{∘} s in θ = 2 1 = s in 3 0 ∘
∴ θ = 30 ∘ ∴ \theta =30^{∘} ∴ θ = 3 0 ∘
θ \theta θ
1 − 2 s i n 2 30 ∘ s i n 30 ∘ = 1 − 2 × 1 2 2 1 2 \frac{1-2sin^{2}30^{∘}}{sin30^{∘}}=\frac{1-2 \times \frac{1}{2}^{2}}{\frac{1}{2}} s in 3 0 ∘ 1 − 2 s i n 2 3 0 ∘ = 2 1 1 − 2 × 2 1 2
= 1 − 2 × 1 4 1 2 = 1 − 1 2 1 2 = 1 2 1 2 = 1 =\frac{1-2 \times \frac{1}{4}}{\frac{1}{2}}=\frac{1-\frac{1}{2}}{\frac{1}{2}}=\frac{\frac{1}{2}}{\frac{1}{2}}=1 = 2 1 1 − 2 × 4 1 = 2 1 1 − 2 1 = 2 1 2 1 = 1
प्रश्न 23. यदि s e c θ + t a n θ = 7 sec \theta +tan \theta =7 sec θ + t an θ = 7 है, तो s e c θ − t a n θ sec \theta -tan \theta sec θ − t an θ का मान लिखिए।
हल— हम जानते हैं। s e c 2 θ − t a n 2 θ = 1 sec^{2} \theta -tan^{2} \theta =1 se c 2 θ − t a n 2 θ = 1
⇒ ( s e c θ + t a n θ ) ( s e c θ − t a n θ ) = 1 ⇒(sec \theta +tan \theta )(sec \theta -tan \theta )=1 ⇒ ( sec θ + t an θ ) ( sec θ − t an θ ) = 1
मान रखने पर
⇒ 7 × ( s e c θ − t a n θ ) = 1 ⇒7 \times (sec \theta -tan \theta )=1 ⇒ 7 × ( sec θ − t an θ ) = 1
∴ ( s e c θ − t a n θ ) = 1 7 ∴(sec \theta -tan \theta )=\frac{1}{7} ∴ ( sec θ − t an θ ) = 7 1
प्रश्न 24. 1 − c o s 2 θ \sqrt{1-cos^{2} \theta } 1 − co s 2 θ का मान θ = 60 ∘ \theta =60^{∘} θ = 6 0 ∘ पर ज्ञात कीजिये। (प्रश्न बैंक; माध्य. शिक्षा बोर्ड, 2023)
हल— 1 − c o s 2 θ \sqrt{1-cos^{2} \theta } 1 − co s 2 θ का मान जब θ = 60 ∘ \theta =60^{∘} θ = 6 0 ∘ हो θ \theta θ का मान रखने पर
1 − c o s 2 60 ∘ = 1 − ( 1 2 ) 2 \sqrt{1-cos^{2}60^{∘}}=\sqrt{1-(\frac{1}{2})^{2}} 1 − co s 2 6 0 ∘ = 1 − ( 2 1 ) 2
= 1 − 1 4 = 3 4 =\sqrt{1-\frac{1}{4}}=\sqrt{\frac{3}{4}} = 1 − 4 1 = 4 3
= 3 2 =\frac{\sqrt{3}}{2} = 2 3
लघूत्तरात्मक प्रश्न—
प्रश्न 1. 4 c o t 2 45 ∘ − s e c 2 60 ∘ + s i n 2 60 ∘ 4cot^{2}45^{∘}-sec^{2}60^{∘}+sin^{2}60^{∘} 4 co t 2 4 5 ∘ − se c 2 6 0 ∘ + s i n 2 6 0 ∘ का मान ज्ञात कीजिए। (माध्य. शिक्षा बोर्ड, 2025)
हल— 4 c o t 2 45 ∘ − s e c 2 60 ∘ + s i n 2 60 ∘ 4cot^{2}45^{∘}-sec^{2}60^{∘}+sin^{2}60^{∘} 4 co t 2 4 5 ∘ − se c 2 6 0 ∘ + s i n 2 6 0 ∘
= 4 ( 1 ) 2 − ( 2 ) 2 + ( 3 2 ) 2 =4(1)^{2}-(2)^{2}+(\frac{\sqrt{3}}{2})^{2} = 4 ( 1 ) 2 − ( 2 ) 2 + ( 2 3 ) 2
= 4 − 4 + 3 4 = 3 4 =4-4+\frac{3}{4}=\frac{3}{4} = 4 − 4 + 4 3 = 4 3
प्रश्न 2. यदि s i n A = 3 5 sinA=\frac{3}{5} s in A = 5 3 हो, तब t a n A + c o s A tanA+cosA t an A + cos A का मान ज्ञात कीजिए। (प्रश्न बैंक; माध्य. शिक्षा बोर्ड, 2022)
हल— माना कि A B C ABC A B C कोई समकोण त्रिभुज है जिसमें कोण B B B पर समकोण है।
[Right-angled triangle ABC with right angle at B, sides AB=4, BC=3, AC=5]
s i n A = 3 5 sinA=\frac{3}{5} s in A = 5 3
( A B ) 2 = ( A C ) 2 − ( B C ) 2 (AB)^{2}=(AC)^{2}-(BC)^{2} ( A B ) 2 = ( A C ) 2 − ( B C ) 2
= ( 5 ) 2 − ( 3 ) 2 =(5)^{2}-(3)^{2} = ( 5 ) 2 − ( 3 ) 2
= 25 − 9 = 16 =25-9=16 = 25 − 9 = 16
∴ A B = 16 = 4 ∴AB=\sqrt{16}=4 ∴ A B = 16 = 4
t a n A = लम्ब आधार = 3 4 tanA=\frac{\text{लम्ब}}{\text{आधार}}=\frac{3}{4} t an A = आधार लम्ब = 4 3
c o s A = आधार कर्ण = 4 5 cosA=\frac{\text{आधार}}{\text{कर्ण}}=\frac{4}{5} cos A = कर्ण आधार = 5 4
तब t a n A + c o s A tanA+cosA t an A + cos A
= 3 4 + 4 5 = 15 + 16 20 = 31 20 =\frac{3}{4}+\frac{4}{5}=\frac{15+16}{20}=\frac{31}{20} = 4 3 + 5 4 = 20 15 + 16 = 20 31
प्रश्न 3. सिद्ध कीजिए: ( s e c A + t a n A ) ( 1 − s i n A ) = c o s A (secA+tanA)(1-sinA)=cosA ( sec A + t an A ) ( 1 − s in A ) = cos A . (माध्य. शिक्षा बोर्ड, 2022; प्रश्न बैंक)
हल— L H S = ( s e c A + t a n A ) ( 1 − s i n A ) LHS=(secA+tanA)(1-sinA) L H S = ( sec A + t an A ) ( 1 − s in A )
⇒ ( 1 c o s A + s i n A c o s A ) ( 1 − s i n A ) ⇒(\frac{1}{cosA}+\frac{sinA}{cosA})(1-sinA) ⇒ ( cos A 1 + cos A s in A ) ( 1 − s in A )
⇒ ( 1 + s i n A ) ( 1 − s i n A ) c o s A = 1 − s i n 2 A c o s A ⇒\frac{(1+sinA)(1-sinA)}{cosA}=\frac{1-sin^{2}A}{cosA} ⇒ cos A ( 1 + s in A ) ( 1 − s in A ) = cos A 1 − s i n 2 A
= c o s 2 A c o s A = c o s A = R H S =\frac{cos^{2}A}{cosA}=cosA=RHS = cos A co s 2 A = cos A = R H S
प्रश्न 4. एक समकोण त्रिभुज A B C ABC A B C में, जिसका कोण B B B समकोण है, यदि t a n A = 1 tanA=1 t an A = 1 तो सत्यापित कीजिए कि 2 s i n A c o s A = 1 2sinAcosA=1 2 s in A cos A = 1 (प्रश्न बैंक)
हल— Δ A B C \Delta ABC Δ A B C में
t a n A = B C A B = 1 tanA=\frac{BC}{AB}=1 t an A = A B B C = 1
अर्थात् B C = A B BC=AB B C = A B
माना A B = B C = K AB=BC=K A B = B C = K
जहाँ पर K K K
पाइथागोरस प्रमेय से
( A C ) 2 = ( A B ) 2 + ( B C ) 2 (AC)^{2}=(AB)^{2}+(BC)^{2} ( A C ) 2 = ( A B ) 2 + ( B C ) 2
= ( K ) 2 + K 2 =(K)^{2}+K^{2} = ( K ) 2 + K 2
[Right-angled triangle ABC with right angle at B]
( A C ) 2 ) (AC)^{2}) ( A C ) 2 )
∴ A C = K 2 ∴AC=K\sqrt{2} ∴ A C = K 2
अतः s i n A = B C A C = K K 2 = 1 2 sinA=\frac{BC}{AC}=\frac{K}{K\sqrt{2}}=\frac{1}{\sqrt{2}} s in A = A C B C = K 2 K = 2 1
और c o s A = A B A C = K K 2 = 1 2 cosA=\frac{AB}{AC}=\frac{K}{K\sqrt{2}}=\frac{1}{\sqrt{2}} cos A = A C A B = K 2 K = 2 1
∴ 2 s i n A c o s A = 2 ( 1 2 ) ( 1 2 ) = 2 2 = 1 ∴2sinAcosA=2(\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}})=\frac{2}{2}=1 ∴ 2 s in A cos A = 2 ( 2 1 ) ( 2 1 ) = 2 2 = 1
प्रश्न 5. Δ P Q R \Delta PQR Δ P QR में, जिसका कोण Q Q Q समकोण है। दी गयी आकृति में, P Q = 3 c m PQ=3cm P Q = 3 c m और P R = 6 c m PR=6cm P R = 6 c m है। ∠ Q P R \angle QPR ∠ QP R और ∠ P R Q \angle PRQ ∠ P R Q ज्ञात कीजिये।
[A right-angled triangle PQR with right angle at Q, PQ=3cm, PR=6cm]
हल— दिया हुआ है—
P Q = 3 c m PQ=3cm P Q = 3 c m P R = 6 c m PR=6cm P R = 6 c m
इसलिये P Q P R = s i n R \frac{PQ}{PR}=sinR P R P Q = s in R
s i n R = 3 6 = 1 2 = s i n 30 ∘ sinR=\frac{3}{6}=\frac{1}{2}=sin30^{∘} s in R = 6 3 = 2 1 = s in 3 0 ∘
अतः ∠ P R Q = 30 ∘ \angle PRQ=30^{∘} ∠ P R Q = 3 0 ∘
और इसलिये ∠ Q P R = 180 ∘ − ( 90 ∘ + 30 ∘ ) = 60 ∘ \angle QPR=180^{∘}-(90^{∘}+30^{∘})=60^{∘} ∠ QP R = 18 0 ∘ − ( 9 0 ∘ + 3 0 ∘ ) = 6 0 ∘
प्रश्न 6. s i n 30 ∘ ⋅ c o s 2 30 ∘ + t a n 45 ∘ ⋅ c o s 2 60 ∘ sin30^{∘}·cos^{2}30^{∘}+tan45^{∘}·cos^{2}60^{∘} s in 3 0 ∘ ⋅ co s 2 3 0 ∘ + t an 4 5 ∘ ⋅ co s 2 6 0 ∘ का मान ज्ञात कीजिये।
हल— s i n 30 ∘ ⋅ c o s 2 30 ∘ + t a n 45 ∘ ⋅ c o s 2 60 ∘ sin30^{∘}·cos^{2}30^{∘}+tan45^{∘}·cos^{2}60^{∘} s in 3 0 ∘ ⋅ co s 2 3 0 ∘ + t an 4 5 ∘ ⋅ co s 2 6 0 ∘
= 1 2 ⋅ 3 2 2 + ( 1 ) 1 2 2 =\frac{1}{2}·\frac{\sqrt{3}}{2}^{2}+(1)\frac{1}{2}^{2} = 2 1 ⋅ 2 3 2 + ( 1 ) 2 1 2
= 1 2 ⋅ 3 4 + ( 1 ) 1 4 = 3 8 + 1 4 =\frac{1}{2}·\frac{3}{4}+(1)\frac{1}{4}=\frac{3}{8}+\frac{1}{4} = 2 1 ⋅ 4 3 + ( 1 ) 4 1 = 8 3 + 4 1
= 3 + 2 8 = 5 8 =\frac{3+2}{8}=\frac{5}{8} = 8 3 + 2 = 8 5
प्रश्न 7. यदि s i n ( A − B ) = 1 2 sin(A-B)=\frac{1}{2} s in ( A − B ) = 2 1 तथा c o s ( A + B ) = 1 2 cos(A+B)=\frac{1}{2} cos ( A + B ) = 2 1 , 0 ∘ < A + B ≤ 90 ∘ 0^{∘}<A+B \le 90^{∘} 0 ∘ < A + B ≤ 9 0 ∘ , A > B A>B A > B , तो A A A और B B B ज्ञात कीजिये। (प्रश्न बैंक)
हल— क्योंकि
s i n ( A − B ) = 1 2 sin(A-B)=\frac{1}{2} s in ( A − B ) = 2 1
∴ s i n ( A − B ) = s i n 30 ∘ ∴sin(A-B)=sin30^{∘} ∴ s in ( A − B ) = s in 3 0 ∘
⇒ A − B = 30 ∘ ⇒A-B=30^{∘} ⇒ A − B = 3 0 ∘
इसी तरह से c o s ( A + B ) = 1 2 cos(A+B)=\frac{1}{2} cos ( A + B ) = 2 1
∴ c o s ( A + B ) = c o s 60 ∘ ∴cos(A+B)=cos60^{∘} ∴ cos ( A + B ) = cos 6 0 ∘
⇒ A + B = 60 ∘ ⇒A+B=60^{∘} ⇒ A + B = 6 0 ∘
समीकरण (i) तथा (ii) को जोड़ने पर
A − B + A + B = 30 ∘ + 60 ∘ = 90 ∘ A-B+A+B=30^{∘}+60^{∘}=90^{∘} A − B + A + B = 3 0 ∘ + 6 0 ∘ = 9 0 ∘
2 A = 90 ∘ 2A=90^{∘} 2 A = 9 0 ∘
A = 45 ∘ A=45^{∘} A = 4 5 ∘
समीकरण (ii) से B = 15 ∘ B=15^{∘} B = 1 5 ∘
प्रश्न 8. ( s e c 2 30 ∘ + c o s e c 2 45 ∘ ) ( 2 c o s 60 ∘ + s i n 90 ∘ + t a n 45 ∘ ) (sec^{2}30^{∘}+cosec^{2}45^{∘})(2cos60^{∘}+sin90^{∘}+tan45^{∘}) ( se c 2 3 0 ∘ + cose c 2 4 5 ∘ ) ( 2 cos 6 0 ∘ + s in 9 0 ∘ + t an 4 5 ∘ ) का मान ज्ञात कीजिये।
हल— ( s e c 2 30 ∘ + c o s e c 2 45 ∘ ) ( 2 c o s 60 ∘ + s i n 90 ∘ + t a n 45 ∘ ) (sec^{2}30^{∘}+cosec^{2}45^{∘})(2cos60^{∘}+sin90^{∘}+tan45^{∘}) ( se c 2 3 0 ∘ + cose c 2 4 5 ∘ ) ( 2 cos 6 0 ∘ + s in 9 0 ∘ + t an 4 5 ∘ )
= 2 3 2 + ( 2 ) 2 2 1 2 + 1 + 1 =\frac{2}{\sqrt{3}}^{2}+(\sqrt{2})^{2}2\frac{1}{2}+1+1 = 3 2 2 + ( 2 ) 2 2 2 1 + 1 + 1
= 4 3 + 2 ( 1 + 1 + 1 ) =\frac{4}{3}+2(1+1+1) = 3 4 + 2 ( 1 + 1 + 1 )
= 4 + 6 3 ( 3 ) = 10 3 × 3 = 10 =\frac{4+6}{3}(3)=\frac{10}{3} \times 3=10 = 3 4 + 6 ( 3 ) = 3 10 × 3 = 10
प्रश्न 9. सिद्ध कीजिये कि (प्रश्न बैंक)
s e c A ( 1 − s i n A ) ( s e c A + t a n A ) = 1 secA(1-sinA)(secA+tanA)=1 sec A ( 1 − s in A ) ( sec A + t an A ) = 1
हल— L.H.S. = s e c A ( 1 − s i n A ) ( s e c A + t a n A ) =secA(1-sinA)(secA+tanA) = sec A ( 1 − s in A ) ( sec A + t an A )
= 1 c o s A ( 1 − s i n A ) 1 c o s A + s i n A c o s A =\frac{1}{cosA}(1-sinA)\frac{1}{cosA}+\frac{sinA}{cosA} = cos A 1 ( 1 − s in A ) cos A 1 + cos A s in A
= ( 1 − s i n A ) ( 1 + s i n A ) c o s 2 A = 1 − s i n 2 A c o s 2 A =\frac{(1-sinA)(1+sinA)}{cos^{2}A}=\frac{1-sin^{2}A}{cos^{2}A} = co s 2 A ( 1 − s in A ) ( 1 + s in A ) = co s 2 A 1 − s i n 2 A
= c o s 2 A c o s 2 A = 1 = R . H . S . =\frac{cos^{2}A}{cos^{2}A}=1=R.H.S. = co s 2 A co s 2 A = 1 = R . H . S .
प्रश्न 10. यदि s i n θ = 1 2 sin \theta =\frac{1}{2} s in θ = 2 1 , तो ( t a n θ + c o t θ ) 2 (tan \theta +cot \theta )^{2} ( t an θ + co tθ ) 2 का मान लिखिए।
हल— ( t a n θ + c o t θ ) 2 = s i n θ c o s θ + c o s θ s i n θ 2 ) (tan \theta +cot \theta )^{2}=\frac{sin \theta }{cos \theta }+\frac{cos \theta }{sin \theta }^{2}) ( t an θ + co tθ ) 2 = cos θ s in θ + s in θ cos θ 2 )
= s i n 2 θ + c o s 2 θ c o s θ ⋅ s i n θ 2 = 1 c o s θ ⋅ s i n θ 2 =\frac{sin^{2} \theta +cos^{2} \theta }{cos \theta ·sin \theta }^{2}=\frac{1}{cos \theta ·sin \theta }^{2} = cos θ ⋅ s in θ s i n 2 θ + co s 2 θ 2 = cos θ ⋅ s in θ 1 2
= 1 c o s 2 θ s i n 2 θ =\frac{1}{cos^{2} \theta sin^{2} \theta } = co s 2 θ s i n 2 θ 1
= 1 s i n 2 θ ( 1 − s i n 2 θ ) = 1 1 2 2 × 1 − 1 2 2 =\frac{1}{sin^{2} \theta (1-sin^{2} \theta )}=\frac{1}{\frac{1}{2}^{2} \times 1-\frac{1}{2}^{2}} = s i n 2 θ ( 1 − s i n 2 θ ) 1 = 2 1 2 × 1 − 2 1 2 1
= 1 1 4 × [ 1 − 1 4 ] = 1 1 4 × 3 4 = 16 3 =\frac{1}{\frac{1}{4} \times [1-\frac{1}{4}]}=\frac{1}{\frac{1}{4} \times \frac{3}{4}}=\frac{16}{3} = 4 1 × [ 1 − 4 1 ] 1 = 4 1 × 4 3 1 = 3 16
प्रश्न 11. यदि s i n A = 3 5 sinA=\frac{3}{5} s in A = 5 3 हो, तो c o s A cosA cos A और c o s e c A cosec A cosec A ज्ञात कीजिए।
हल— माना कि A B C ABC A B C कोई समकोण त्रिभुज है जिसमें कोण B B B पर समकोण है। ∵ s i n A = 3 5 ∵sinA=\frac{3}{5} ∵ s in A = 5 3
[Right angled triangle ABC with angle A at vertex A, angle B=90 degrees, BC=3k, AC=5k, AB=4k]
परन्तु s i n A = लम्ब कर्ण = B C A C sinA=\frac{\text{लम्ब}}{\text{कर्ण}}=\frac{BC}{AC} s in A = कर्ण लम्ब = A C B C
∴ B C A C = 3 5 ∴\frac{BC}{AC}=\frac{3}{5} ∴ A C B C = 5 3
माना B C = 3 k BC=3k B C = 3 k
A C = 5 k AC=5k A C = 5 k
पाइथागोरस प्रमेय से
A C 2 = A B 2 + B C 2 AC^{2}=AB^{2}+BC^{2} A C 2 = A B 2 + B C 2
या ( 5 k ) 2 = ( A B ) 2 + ( 3 k ) 2 (5k)^{2}=(AB)^{2}+(3k)^{2} ( 5 k ) 2 = ( A B ) 2 + ( 3 k ) 2
या 25 k 2 = A B 2 + 9 k 2 25k^{2}=AB^{2}+9k^{2} 25 k 2 = A B 2 + 9 k 2
या 25 k 2 − 9 k 2 = A B 2 25k^{2}-9k^{2}=AB^{2} 25 k 2 − 9 k 2 = A B 2
या 16 k 2 = A B 2 16k^{2}=AB^{2} 16 k 2 = A B 2
या A B = 4 k AB=4k A B = 4 k
∴ c o s A = आधार कर्ण = A B A C = 4 5 ∴cosA=\frac{\text{आधार}}{\text{कर्ण}}=\frac{AB}{AC}=\frac{4}{5} ∴ cos A = कर्ण आधार = A C A B = 5 4
तथा c o s e c A = 1 s i n A = कर्ण लम्ब = A C B C cosec A=\frac{1}{sinA}=\frac{\text{कर्ण}}{\text{लम्ब}}=\frac{AC}{BC} cosec A = s in A 1 = लम्ब कर्ण = B C A C
या c o s e c A = 5 3 cosec A=\frac{5}{3} cosec A = 3 5
प्रश्न 12. सिद्ध कीजिए कि 1 − t a n A 1 − c o t A 2 = t a n 2 A \frac{1-tanA}{1-cotA}^{2}=tan^{2}A 1 − co t A 1 − t an A 2 = t a n 2 A (प्रश्न बैंक)
हल— L.H.S. = 1 − t a n A 1 − c o t A 2 =\frac{1-tanA}{1-cotA}^{2} = 1 − co t A 1 − t an A 2
= 1 − s i n A c o s A 1 − c o s A s i n A 2 = c o s A − s i n A c o s A s i n A − c o s A s i n A 2 =\frac{1-\frac{sinA}{cosA}}{1-\frac{cosA}{sinA}}^{2}=\frac{\frac{cosA-sinA}{cosA}}{\frac{sinA-cosA}{sinA}}^{2} = 1 − s in A cos A 1 − cos A s in A 2 = s in A s in A − cos A cos A cos A − s in A 2
= − ( s i n A − c o s A ) c o s A × s i n A ( s i n A ) 2 =\frac{-(sinA-cosA)}{cosA} \times \frac{sinA}{(sinA)}^{2} = cos A − ( s in A − cos A ) × ( s in A ) s in A 2
= − s i n A c o s A 2 = s i n 2 A c o s 2 A =-\frac{sinA}{cosA}^{2}=\frac{sin^{2}A}{cos^{2}A} = − cos A s in A 2 = co s 2 A s i n 2 A
= t a n 2 A =tan^{2}A = t a n 2 A
= R . H . S . =R.H.S. = R . H . S .
∴ L . H . S . = R . H . S . ∴L.H.S.=R.H.S. ∴ L . H . S . = R . H . S .
प्रश्न 13. c o s 45 ∘ s e c 30 ∘ + c o s e c 30 ∘ \frac{cos45^{∘}}{sec30^{∘}+cosec 30^{∘}} sec 3 0 ∘ + cosec 3 0 ∘ cos 4 5 ∘ का मान ज्ञात कीजिये।
हल— c o s 45 ∘ s e c 30 ∘ + c o s e c 30 ∘ \frac{cos45^{∘}}{sec30^{∘}+cosec 30^{∘}} sec 3 0 ∘ + cosec 3 0 ∘ cos 4 5 ∘
= 1 2 2 3 + 2 = 1 2 2 + 2 3 3 =\frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}}+2}=\frac{\frac{1}{\sqrt{2}}}{\frac{2+2\sqrt{3}}{\sqrt{3}}} = 3 2 + 2 2 1 = 3 2 + 2 3 2 1
⇒ 1 2 ⋅ 3 2 + 2 3 = 3 2 2 ( 1 + 3 ) ⇒\frac{1}{\sqrt{2}}·\frac{\sqrt{3}}{2+2\sqrt{3}}=\frac{\sqrt{3}}{2\sqrt{2}(1+\sqrt{3})} ⇒ 2 1 ⋅ 2 + 2 3 3 = 2 2 ( 1 + 3 ) 3
= 6 4 3 − 1 ( 3 + 1 ) ( 3 − 1 ) =\frac{\sqrt{6}}{4}\frac{\sqrt{3}-1}{(\sqrt{3}+1)(\sqrt{3}-1)} = 4 6 ( 3 + 1 ) ( 3 − 1 ) 3 − 1
= 6 ( 3 − 1 ) 4 ( 3 − 1 ) = 6 ( 3 − 1 ) 8 =\frac{\sqrt{6}(\sqrt{3}-1)}{4(3-1)}=\frac{\sqrt{6}(\sqrt{3}-1)}{8} = 4 ( 3 − 1 ) 6 ( 3 − 1 ) = 8 6 ( 3 − 1 )
प्रश्न 14. सिद्ध कीजिए :
3 t a n 2 30 ∘ − 4 3 s i n 2 60 ∘ − 1 2 c o s e c 2 45 ∘ + 4 3 s i n 2 90 ∘ = 1 3 3tan^{2}30^{∘}-\frac{4}{3}sin^{2}60^{∘}-\frac{1}{2}cosec^{2}45^{∘}+\frac{4}{3}sin^{2}90^{∘}=\frac{1}{3} 3 t a n 2 3 0 ∘ − 3 4 s i n 2 6 0 ∘ − 2 1 cose c 2 4 5 ∘ + 3 4 s i n 2 9 0 ∘ = 3 1
हल— बायाँ पक्ष (L.H.S.)
3 t a n 2 30 ∘ − 4 3 s i n 2 60 ∘ − 1 2 c o s e c 2 45 ∘ + 4 3 s i n 2 90 ∘ 3tan^{2}30^{∘}-\frac{4}{3}sin^{2}60^{∘}-\frac{1}{2}cosec^{2}45^{∘}+\frac{4}{3}sin^{2}90^{∘} 3 t a n 2 3 0 ∘ − 3 4 s i n 2 6 0 ∘ − 2 1 cose c 2 4 5 ∘ + 3 4 s i n 2 9 0 ∘
= 3 1 3 2 − 4 3 3 2 2 − 1 2 ( 2 ) 2 + 4 3 ( 1 ) 2 =3\frac{1}{\sqrt{3}}^{2}-\frac{4}{3}\frac{\sqrt{3}}{2}^{2}-\frac{1}{2}(\sqrt{2})^{2}+\frac{4}{3}(1)^{2} = 3 3 1 2 − 3 4 2 3 2 − 2 1 ( 2 ) 2 + 3 4 ( 1 ) 2
= 3 1 3 − 4 3 3 4 − 1 2 ( 2 ) + 4 3 ( 1 ) =3\frac{1}{3}-\frac{4}{3}\frac{3}{4}-\frac{1}{2}(2)+\frac{4}{3}(1) = 3 3 1 − 3 4 4 3 − 2 1 ( 2 ) + 3 4 ( 1 )
= 1 − 1 − 1 + 4 3 = 1 3 =1-1-1+\frac{4}{3}=\frac{1}{3} = 1 − 1 − 1 + 3 4 = 3 1
प्रश्न 15. किसी त्रिभुज A B C ABC A B C में A B = 24 AB=24 A B = 24 सेमी, B C = 7 BC=7 B C = 7 सेमी तथा ∠ B = 90 ∘ \angle B=90^{∘} ∠ B = 9 0 ∘ है तो s i n A sinA s in A व s i n C sinC s in C का मान ज्ञात कीजिए।
(माध्य. शिक्षा बोर्ड, मॉडल पेपर, 2023-24)
हल— ज्ञात करना है s i n A , s i n C sinA,sinC s in A , s in C
यहाँ A B = 24 AB=24 A B = 24 B C = 7 BC=7 B C = 7 ∠ B = 90 ∘ \angle B=90^{∘} ∠ B = 9 0 ∘
पाइथागोरस प्रमेय से
A C 2 = A B 2 + B C 2 AC^{2}=AB^{2}+BC^{2} A C 2 = A B 2 + B C 2
A C 2 = ( 24 ) 2 + ( 7 ) 2 AC^{2}=(24)^{2}+(7)^{2} A C 2 = ( 24 ) 2 + ( 7 ) 2
A C 2 = 576 + 49 AC^{2}=576+49 A C 2 = 576 + 49
A C 2 = 625 AC^{2}=625 A C 2 = 625
A C = 625 AC=\sqrt{625} A C = 625
A C = 25 AC=25 A C = 25
[Right-angled triangle ABC with AB=24cm, BC=7cm, AC=25cm]
s i n A = B C A C sinA=\frac{BC}{AC} s in A = A C B C
∴ s i n A = 7 c m 25 c m = 7 25 ∴sinA=\frac{7 cm}{25 cm}=\frac{7}{25} ∴ s in A = 25 c m 7 c m = 25 7 उत्तर
s i n C = ∠ C कीसम्मुखभुजा कर्ण sinC=\frac{ \angle C \text{की} \text{सम्मुख} \text{भुजा}}{\text{कर्ण}} s in C = कर्ण ∠ C की सम्मुख भुजा
= A B A C = 24 c m 25 c m =\frac{AB}{AC}=\frac{24 cm}{25 cm} = A C A B = 25 c m 24 c m
∴ s i n C = 24 25 ∴sinC=\frac{24}{25} ∴ s in C = 25 24 उत्तर
प्रश्न 16. यदि t a n A = 1 tanA=1 t an A = 1 हो, तो 2 s i n A c o s A 2sinAcosA 2 s in A cos A का मान ज्ञात कीजिए।
(माध्य. शिक्षा बोर्ड, 2024)
हल— दिया है—
t a n A = 1 tanA=1 t an A = 1
∴ t a n A = t a n 45 ∘ ∴tanA=tan45^{∘} ∴ t an A = t an 4 5 ∘
∴ A = 45 ∘ ∴A=45^{∘} ∴ A = 4 5 ∘
तब 2 s i n 45 ∘ c o s 45 ∘ = 2 × 1 2 × 1 2 2sin45^{∘}cos45^{∘}=2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} 2 s in 4 5 ∘ cos 4 5 ∘ = 2 × 2 1 × 2 1
= 2 2 × 2 =\frac{2}{\sqrt{2} \times \sqrt{2}} = 2 × 2 2
= 2 2 = 1 =\frac{2}{2}=1 = 2 2 = 1 उत्तर
दीर्घउत्तरीय प्रश्न—
प्रश्न 1. Δ O P Q \Delta OPQ Δ O P Q में, जिसका कोण P P P समकोण है, O P = 7 OP=7 O P = 7 सेमी और O Q − P Q = 1 OQ-PQ=1 O Q − P Q = 1 सेमी (देखिये आकृति), s i n Q sinQ s in Q और c o s Q cosQ cos Q के मान ज्ञात कीजिए।
हल— चित्र की आकृति से
O Q 2 = O P 2 + P Q 2 OQ^{2}=OP^{2}+PQ^{2} O Q 2 = O P 2 + P Q 2
[Right-angled triangle OPQ with P=90 degrees, OP=7cm]
लेकिन दिया है
O Q − P Q = 1 OQ-PQ=1 O Q − P Q = 1
∴ O Q = 1 + P Q ∴OQ=1+PQ ∴ O Q = 1 + P Q
अर्थात् ( 1 + P Q ) 2 = O P 2 + P Q 2 ) (1+PQ)^{2}=OP^{2}+PQ^{2}) ( 1 + P Q ) 2 = O P 2 + P Q 2 )
अर्थात् 1 + P Q 2 + 2 P Q = O P 2 + P Q 2 1+PQ^{2}+2PQ=OP^{2}+PQ^{2} 1 + P Q 2 + 2 P Q = O P 2 + P Q 2
अर्थात् 1 + 2 P Q = ( 7 ) 2 1+2PQ=(7)^{2} 1 + 2 P Q = ( 7 ) 2
1 + 2 P Q = 49 1+2PQ=49 1 + 2 P Q = 49
∴ P Q = 24 ∴PQ=24 ∴ P Q = 24
और O Q = 1 + P Q = 1 + 24 = 25 OQ=1+PQ=1+24=25 O Q = 1 + P Q = 1 + 24 = 25
s i n Q = O P O Q = 7 25 sinQ=\frac{OP}{OQ}=\frac{7}{25} s in Q = O Q O P = 25 7
और c o s Q = P Q O Q = 24 25 cosQ=\frac{PQ}{OQ}=\frac{24}{25} cos Q = O Q P Q = 25 24 उत्तर
प्रश्न 2. यदि t a n A = 3 4 tanA=\frac{3}{4} t an A = 4 3 हो, तो s e c A ( 1 − s i n A ) ( s e c A + t a n A ) secA(1-sinA)(secA+tanA) sec A ( 1 − s in A ) ( sec A + t an A ) का मान ज्ञात कीजिए।
हल— माना कि A B C ABC A B C एक समकोण त्रिभुज है, जिसमें कोण B B B समकोण है।
∠ A \angle A ∠ A
[Right-angled triangle ABC with AB=4k, BC=3k, AC=5k]
आधार = A B =AB = A B = B C =BC = B C = A C =AC = A C
दिया गया है, t a n A = 3 4 tanA=\frac{3}{4} t an A = 4 3
परन्तु t a n A = लम्ब आधार = B C A B tanA=\frac{\text{लम्ब}}{\text{आधार}}=\frac{BC}{AB} t an A = आधार लम्ब = A B B C
B C A B = 3 4 = k \frac{BC}{AB}=\frac{3}{4}=k A B B C = 4 3 = k
∴ A B = 4 k , B C = 3 k ∴AB=4k,BC=3k ∴ A B = 4 k , B C = 3 k
समकोण Δ A B C \Delta ABC Δ A B C
( A C ) 2 = ( A B ) 2 + ( B C ) 2 (AC)^{2}=(AB)^{2}+(BC)^{2} ( A C ) 2 = ( A B ) 2 + ( B C ) 2
= ( 4 k ) 2 + ( 3 k ) 2 =(4k)^{2}+(3k)^{2} = ( 4 k ) 2 + ( 3 k ) 2
( A C ) 2 = 16 k 2 + 9 k 2 = 25 k 2 ) (AC)^{2}=16k^{2}+9k^{2}=25k^{2}) ( A C ) 2 = 16 k 2 + 9 k 2 = 25 k 2 )
∴ A C = ± 25 k 2 = ± 5 k ∴AC= \pm \sqrt{25k^{2}}= \pm 5k ∴ A C = ± 25 k 2 = ± 5 k
A C = 5 k AC=5k A C = 5 k
( ∵ A C ≠ − 5 k ) (∵AC \ne -5k) ( ∵ A C = − 5 k ) ] ] ]
s i n A = लम्ब कर्ण = B C A C = 3 k 5 k sinA=\frac{\text{लम्ब}}{\text{कर्ण}}=\frac{BC}{AC}=\frac{3k}{5k} s in A = कर्ण लम्ब = A C B C = 5 k 3 k
s i n A = 3 5 sinA=\frac{3}{5} s in A = 5 3
s e c A = कर्ण आधार = A C A B = 5 k 4 k secA=\frac{\text{कर्ण}}{\text{आधार}}=\frac{AC}{AB}=\frac{5k}{4k} sec A = आधार कर्ण = A B A C = 4 k 5 k
s e c A = 5 4 secA=\frac{5}{4} sec A = 4 5
और t a n A = 3 4 tanA=\frac{3}{4} t an A = 4 3
∴ s e c A ( 1 − s i n A ) ( s e c A + t a n A ) ∴secA(1-sinA)(secA+tanA) ∴ sec A ( 1 − s in A ) ( sec A + t an A )
= 5 4 ( 1 − 3 5 ) ( 5 4 + 3 4 ) =\frac{5}{4}(1-\frac{3}{5})(\frac{5}{4}+\frac{3}{4}) = 4 5 ( 1 − 5 3 ) ( 4 5 + 4 3 )
= 5 4 × 2 5 × 8 4 = 1 =\frac{5}{4} \times \frac{2}{5} \times \frac{8}{4}=1 = 4 5 × 5 2 × 4 8 = 1
प्रश्न 3. यदि 3 c o t A = 4 3cotA=4 3 co t A = 4 , तो 1 − t a n 2 A 1 + t a n 2 A \frac{1-tan^{2}A}{1+tan^{2}A} 1 + t a n 2 A 1 − t a n 2 A का मान ज्ञात कीजिए। (प्रश्न बैंक)
हल— ∵ 3 c o t A = 4 ∴ c o t A = 4 3 ∵3cotA=4∴cotA=\frac{4}{3} ∵ 3 co t A = 4 ∴ co t A = 3 4
हम जानते हैं कि
t a n A = 1 c o t A tanA=\frac{1}{cotA} t an A = co t A 1
∴ t a n A = 1 4 3 = 3 4 ∴tanA=\frac{1}{\frac{4}{3}}=\frac{3}{4} ∴ t an A = 3 4 1 = 4 3
अब प्रश्नानुसार
1 − t a n 2 A 1 + t a n 2 A = 1 − ( 3 4 ) 2 1 + ( 3 4 ) 2 \frac{1-tan^{2}A}{1+tan^{2}A}=\frac{1-(\frac{3}{4})^{2}}{1+(\frac{3}{4})^{2}} 1 + t a n 2 A 1 − t a n 2 A = 1 + ( 4 3 ) 2 1 − ( 4 3 ) 2
= 1 − 9 16 1 + 9 16 = 16 − 9 16 16 + 9 16 = 7 16 25 16 =\frac{1-\frac{9}{16}}{1+\frac{9}{16}}=\frac{\frac{16-9}{16}}{\frac{16+9}{16}}=\frac{\frac{7}{16}}{\frac{25}{16}} = 1 + 16 9 1 − 16 9 = 16 16 + 9 16 16 − 9 = 16 25 16 7
= 7 16 × 16 25 = 7 25 =\frac{7}{16} \times \frac{16}{25}=\frac{7}{25} = 16 7 × 25 16 = 25 7
प्रश्न 4. यदि c o s A = 12 13 cosA=\frac{12}{13} cos A = 13 12 तो c o t A cotA co t A का मान परिकलित कीजिए।
हल— माना कि A B C ABC A B C कोई समकोण त्रिभुज है जिसमें कोण B B B पर समकोण है।
[Right-angled triangle ABC with right angle at B, hypotenuse AC=13k, base AB=12k, perpendicular BC=5k]
c o s A = 12 13 cosA=\frac{12}{13} cos A = 13 12
परन्तु c o s A = आधार कर्ण = A B A C cosA=\frac{\text{आधार}}{\text{कर्ण}}=\frac{AB}{AC} cos A = कर्ण आधार = A C A B
∴ A B A C = 12 13 ∴\frac{AB}{AC}=\frac{12}{13} ∴ A C A B = 13 12
माना A B = 12 k , A C = 13 k AB=12k,AC=13k A B = 12 k , A C = 13 k
पाइथागोरस प्रमेय से
A C 2 = A B 2 + B C 2 AC^{2}=AB^{2}+BC^{2} A C 2 = A B 2 + B C 2
∴ B C 2 = A C 2 − A B 2 ∴BC^{2}=AC^{2}-AB^{2} ∴ B C 2 = A C 2 − A B 2
= ( 13 k ) 2 − ( 12 k ) 2 =(13k)^{2}-(12k)^{2} = ( 13 k ) 2 − ( 12 k ) 2
= 169 k 2 − 144 k 2 = 25 k 2 =169k^{2}-144k^{2}=25k^{2} = 169 k 2 − 144 k 2 = 25 k 2
∴ B C = 5 k ∴BC=5k ∴ B C = 5 k
इसलिए c o t A = आधार लम्ब = 12 k 5 k = 12 5 cotA=\frac{\text{आधार}}{\text{लम्ब}}=\frac{12k}{5k}=\frac{12}{5} co t A = लम्ब आधार = 5 k 12 k = 5 12
प्रश्न 5. यदि s i n ( A + B ) = 1 sin(A+B)=1 s in ( A + B ) = 1 तथा c o s ( A − B ) = 3 2 cos(A-B)=\frac{\sqrt{3}}{2} cos ( A − B ) = 2 3 जहाँ 0 ∘ B 0^{∘} B 0 ∘ B हो, तो A A A तथा B B B के मान ज्ञात कीजिए।
हल— दिया है।
s i n ( A + B ) = 1 sin(A+B)=1 s in ( A + B ) = 1
⇒ ∴ s i n ( A + B ) = s i n 90 ∘ ⇒∴sin(A+B)=sin90^{∘} ⇒ ∴ s in ( A + B ) = s in 9 0 ∘
या A + B = 90 ∘ A+B=90^{∘} A + B = 9 0 ∘
तथा c o s ( A − B ) = 3 2 cos(A-B)=\frac{\sqrt{3}}{2} cos ( A − B ) = 2 3
या c o s ( A − B ) = c o s 30 ∘ cos(A-B)=cos30^{∘} cos ( A − B ) = cos 3 0 ∘
या A − B = 30 ∘ A-B=30^{∘} A − B = 3 0 ∘
समी. (1) तथा (2) को जोड़ने पर
A + B + A − B = 90 ∘ + 30 ∘ A+B+A-B=90^{∘}+30^{∘} A + B + A − B = 9 0 ∘ + 3 0 ∘
2 A = 120 ∘ ∴ A = 60 ∘ 2A=120^{∘}∴A=60^{∘} 2 A = 12 0 ∘ ∴ A = 6 0 ∘
A A A
60 ∘ + B = 90 ∘ 60^{∘}+B=90^{∘} 6 0 ∘ + B = 9 0 ∘
B = 90 ∘ − 60 ∘ = 30 ∘ B=90^{∘}-60^{∘}=30^{∘} B = 9 0 ∘ − 6 0 ∘ = 3 0 ∘
∴ A = 60 ∘ , B = 30 ∘ ∴A=60^{∘},B=30^{∘} ∴ A = 6 0 ∘ , B = 3 0 ∘
प्रश्न 6. यदि t a n A = 2 − 1 tanA=\sqrt{2}-1 t an A = 2 − 1 हो, तो सिद्ध करो कि s i n A c o s A = 2 4 sinAcosA=\frac{\sqrt{2}}{4} s in A cos A = 4 2
हल— t a n A = लम्ब आधार = 2 − 1 1 tanA=\frac{\text{लम्ब}}{\text{आधार}}=\frac{\sqrt{2}-1}{1} t an A = आधार लम्ब = 1 2 − 1
त्रिभुज A B C ABC A B C
A C 2 = A B 2 + B C 2 AC^{2}=AB^{2}+BC^{2} A C 2 = A B 2 + B C 2
A C 2 = ( 1 ) 2 + ( 2 − 1 ) 2 AC^{2}=(1)^{2}+(\sqrt{2}-1)^{2} A C 2 = ( 1 ) 2 + ( 2 − 1 ) 2
= 1 + 2 + 1 − 2 2 =1+2+1-2\sqrt{2} = 1 + 2 + 1 − 2 2
= 4 − 2 2 =4-2\sqrt{2} = 4 − 2 2
A C = 4 − 2 2 AC=\sqrt{4-2\sqrt{2}} A C = 4 − 2 2
[Right angled triangle ABC with base AB=1, height BC=√(2)-1 and hypotenuse AC=√(4-2√2)]
s i n A = B C A C = 2 − 1 4 − 2 2 sinA=\frac{BC}{AC}=\frac{\sqrt{2}-1}{\sqrt{4-2\sqrt{2}}} s in A = A C B C = 4 − 2 2 2 − 1
और c o s A = A B A C = 1 4 − 2 2 cosA=\frac{AB}{AC}=\frac{1}{\sqrt{4-2\sqrt{2}}} cos A = A C A B = 4 − 2 2 1
अतः s i n A c o s A = 2 − 1 4 − 2 2 × 1 4 − 2 2 sinAcosA=\frac{\sqrt{2}-1}{\sqrt{4-2\sqrt{2}}} \times \frac{1}{\sqrt{4-2\sqrt{2}}} s in A cos A = 4 − 2 2 2 − 1 × 4 − 2 2 1
= 2 − 1 4 − 2 2 = 2 − 1 2 2 ( 2 − 1 ) =\frac{\sqrt{2}-1}{4-2\sqrt{2}}=\frac{\sqrt{2}-1}{2\sqrt{2}(\sqrt{2}-1)} = 4 − 2 2 2 − 1 = 2 2 ( 2 − 1 ) 2 − 1
= 1 2 2 = 2 4 =\frac{1}{2\sqrt{2}}=\frac{\sqrt{2}}{4} = 2 2 1 = 4 2
प्रश्न 7. यदि c o t B = 12 5 cotB=\frac{12}{5} co tB = 5 12 हो, तो सिद्ध करो कि t a n 2 B − s i n 2 B = s i n 4 B s e c 2 B tan^{2}B-sin^{2}B=sin^{4}Bsec^{2}B t a n 2 B − s i n 2 B = s i n 4 B se c 2 B
हल— दिया है,
c o t B = आधार लम्ब = 12 5 cotB=\frac{\text{आधार}}{\text{लम्ब}}=\frac{12}{5} co tB = लम्ब आधार = 5 12
पाइथागोरस प्रमेय से
( A B ) 2 = ( B C ) 2 + ( A C ) 2 (AB)^{2}=(BC)^{2}+(AC)^{2} ( A B ) 2 = ( B C ) 2 + ( A C ) 2
= ( 12 ) 2 + ( 5 ) 2 =(12)^{2}+(5)^{2} = ( 12 ) 2 + ( 5 ) 2
= 144 + 25 = 169 =144+25=169 = 144 + 25 = 169
∴ A B = 13 ∴AB=13 ∴ A B = 13
[Right angled triangle ABC with base BC=12, height AC=5 and hypotenuse AB=13]
s i n B = A C A B = 5 13 sinB=\frac{AC}{AB}=\frac{5}{13} s in B = A B A C = 13 5
t a n B = A C A B = 5 12 tanB=\frac{AC}{AB}=\frac{5}{12} t an B = A B A C = 12 5 s e c B = A B B C = 13 12 secB=\frac{AB}{BC}=\frac{13}{12} sec B = B C A B = 12 13
L H S = t a n 2 B − s i n 2 B LHS=tan^{2}B-sin^{2}B L H S = t a n 2 B − s i n 2 B
= 5 12 2 − 5 13 2 = 25 144 − 25 169 =\frac{5}{12}^{2}-\frac{5}{13}^{2}=\frac{25}{144}-\frac{25}{169} = 12 5 2 − 13 5 2 = 144 25 − 169 25
= 25 169 − 144 144 × 169 = 25 × 25 144 × 169 =25\frac{169-144}{144 \times 169}=\frac{25 \times 25}{144 \times 169} = 25 144 × 169 169 − 144 = 144 × 169 25 × 25
L H S = 5 2 × 5 2 ( 12 ) 2 × ( 13 ) 2 LHS=\frac{5^{2} \times 5^{2}}{(12)^{2} \times (13)^{2}} L H S = ( 12 ) 2 × ( 13 ) 2 5 2 × 5 2
R . H . S . = s i n 4 B s e c 2 B = 5 13 4 × 13 12 2 R.H.S.=sin^{4}Bsec^{2}B=\frac{5}{13}^{4} \times \frac{13}{12}^{2} R . H . S . = s i n 4 B se c 2 B = 13 5 4 × 12 13 2
= 5 4 × 13 2 13 4 × ( 12 ) 2 = 5 2 × 5 2 ( 13 ) 2 × ( 12 ) 2 = L . H . S . =\frac{5^{4} \times 13^{2}}{13^{4} \times (12)^{2}}=\frac{5^{2} \times 5^{2}}{(13)^{2} \times (12)^{2}}=L.H.S. = 1 3 4 × ( 12 ) 2 5 4 × 1 3 2 = ( 13 ) 2 × ( 12 ) 2 5 2 × 5 2 = L . H . S .
निबन्धात्मक प्रश्न—
प्रश्न 1. सिद्ध कीजिए कि—
1 1 + s i n θ + 1 1 − s i n θ = 2 s e c 2 θ \frac{1}{1+sin \theta }+\frac{1}{1-sin \theta }=2sec^{2} \theta 1 + s in θ 1 + 1 − s in θ 1 = 2 se c 2 θ
हल— L H S = 1 1 + s i n θ + 1 1 − s i n θ LHS=\frac{1}{1+sin \theta }+\frac{1}{1-sin \theta } L H S = 1 + s in θ 1 + 1 − s in θ 1
= ( 1 − s i n θ ) + ( 1 + s i n θ ) ( 1 + s i n θ ) ( 1 − s i n θ ) =\frac{(1-sin \theta )+(1+sin \theta )}{(1+sin \theta )(1-sin \theta )} = ( 1 + s in θ ) ( 1 − s in θ ) ( 1 − s in θ ) + ( 1 + s in θ )
= 2 1 − s i n 2 θ = 2 c o s 2 θ =\frac{2}{1-sin^{2} \theta }=\frac{2}{cos^{2} \theta } = 1 − s i n 2 θ 2 = co s 2 θ 2
= 2 s e c 2 θ = R H S =2sec^{2} \theta =RHS = 2 se c 2 θ = R H S
प्रश्न 2. सिद्ध कीजिए कि—
s i n 2 θ c o s θ + c o s 3 θ + t a n θ s i n θ = s e c θ sin^{2} \theta cos \theta +cos^{3} \theta +tan \theta sin \theta =sec \theta s i n 2 θ cos θ + co s 3 θ + t an θ s in θ = sec θ
हल— L H S = s i n 2 θ c o s θ + c o s 3 θ + t a n θ s i n θ LHS=sin^{2} \theta cos \theta +cos^{3} \theta +tan \theta sin \theta L H S = s i n 2 θ cos θ + co s 3 θ + t an θ s in θ
= c o s θ ( s i n 2 θ + c o s 2 θ ) + s i n θ c o s θ ⋅ s i n θ =cos \theta (sin^{2} \theta +cos^{2} \theta )+\frac{sin \theta }{cos \theta }·sin \theta = cos θ ( s i n 2 θ + co s 2 θ ) + cos θ s in θ ⋅ s in θ
= c o s θ + s i n 2 θ c o s θ =cos \theta +\frac{sin^{2} \theta }{cos \theta } = cos θ + cos θ s i n 2 θ
= c o s 2 θ + s i n 2 θ c o s θ = 1 c o s θ =\frac{cos^{2} \theta +sin^{2} \theta }{cos \theta }=\frac{1}{cos \theta } = cos θ co s 2 θ + s i n 2 θ = cos θ 1
= s e c θ = R H S =sec \theta =RHS = sec θ = R H S
प्रश्न 3. सिद्ध कीजिए-
( माध्य. शिक्षा बोर्ड, मॉडल पेपर, 2024-25 )
s i n 6 A + c o s 6 A = 1 − 3 s i n 2 A c o s 2 A sin^{6}A+cos^{6}A=1-3sin^{2}Acos^{2}A s i n 6 A + co s 6 A = 1 − 3 s i n 2 A co s 2 A
हल—
L H S = s i n 6 A + c o s 6 A LHS=sin^{6}A+cos^{6}A L H S = s i n 6 A + co s 6 A
= ( s i n 2 A ) 3 + ( c o s 2 A ) 3 =(sin^{2}A)^{3}+(cos^{2}A)^{3} = ( s i n 2 A ) 3 + ( co s 2 A ) 3
= ( s i n 2 A + c o s 2 A ) [ ( s i n 2 A ) 2 + ( c o s 2 A ) 2 − s i n 2 A c o s 2 A ] =(sin^{2}A+cos^{2}A)[(sin^{2}A)^{2}+(cos^{2}A)^{2}-sin^{2}Acos^{2}A] = ( s i n 2 A + co s 2 A ) [( s i n 2 A ) 2 + ( co s 2 A ) 2 − s i n 2 A co s 2 A ]
( ∵ a 3 + b 3 = ( a + b ) ( a 2 + b 2 − a b ) ) (∵a^{3}+b^{3}=(a+b)(a^{2}+b^{2}-ab)) ( ∵ a 3 + b 3 = ( a + b ) ( a 2 + b 2 − ab ))
= ( 1 ) ( s i n 2 A ) 2 + ( c o s 2 A ) 2 + 2 s i n 2 A c o s 2 A − 2 s i n 2 A c o s 2 A − s i n 2 A c o s 2 A =(1){(sin^{2}A)^{2}+(cos^{2}A)^{2}+2sin^{2}Acos^{2}A-2sin^{2}Acos^{2}A-sin^{2}Acos^{2}A} = ( 1 ) ( s i n 2 A ) 2 + ( co s 2 A ) 2 + 2 s i n 2 A co s 2 A − 2 s i n 2 A co s 2 A − s i n 2 A co s 2 A
= ( s i n 4 A + c o s 4 A + 2 s i n 2 A c o s 2 A − 3 s i n 2 A c o s 2 A ) =(sin^{4}A+cos^{4}A+2sin^{2}Acos^{2}A-3sin^{2}Acos^{2}A) = ( s i n 4 A + co s 4 A + 2 s i n 2 A co s 2 A − 3 s i n 2 A co s 2 A )
= ( s i n 2 A + c o s 2 A ) 2 − 3 s i n 2 A c o s 2 A =(sin^{2}A+cos^{2}A)^{2}-3sin^{2}Acos^{2}A = ( s i n 2 A + co s 2 A ) 2 − 3 s i n 2 A co s 2 A
( ∵ ( a + b ) 2 = a 2 + b 2 + 2 a b ) (∵(a+b)^{2}=a^{2}+b^{2}+2ab) ( ∵ ( a + b ) 2 = a 2 + b 2 + 2 ab )
= ( 1 ) 2 − 3 s i n 2 A c o s 2 A =(1)^{2}-3sin^{2}Acos^{2}A = ( 1 ) 2 − 3 s i n 2 A co s 2 A
= 1 − 3 s i n 2 A c o s 2 A =1-3sin^{2}Acos^{2}A = 1 − 3 s i n 2 A co s 2 A
= R H S =RHS = R H S
प्रश्न 4. सिद्ध कीजिए-
c o s 4 θ + s i n 4 θ = 1 − 2 c o s 2 θ s i n 2 θ cos^{4} \theta +sin^{4} \theta =1-2cos^{2} \theta sin^{2} \theta co s 4 θ + s i n 4 θ = 1 − 2 co s 2 θ s i n 2 θ
( माध्य. शिक्षा बोर्ड, मॉडल पेपर, 2021-22 )
हल—
L H S = c o s 4 θ + s i n 4 θ LHS=cos^{4} \theta +sin^{4} \theta L H S = co s 4 θ + s i n 4 θ
= ( c o s 2 θ ) 2 + ( s i n 2 θ ) 2 + 2 c o s 2 θ s i n 2 θ − 2 c o s 2 θ s i n 2 θ =(cos^{2} \theta )^{2}+(sin^{2} \theta )^{2}+2cos^{2} \theta sin^{2} \theta -2cos^{2} \theta sin^{2} \theta = ( co s 2 θ ) 2 + ( s i n 2 θ ) 2 + 2 co s 2 θ s i n 2 θ − 2 co s 2 θ s i n 2 θ
= ( c o s 2 θ + s i n 2 θ ) 2 − 2 c o s 2 θ s i n 2 θ =(cos^{2} \theta +sin^{2} \theta )^{2}-2cos^{2} \theta sin^{2} \theta = ( co s 2 θ + s i n 2 θ ) 2 − 2 co s 2 θ s i n 2 θ
= ( 1 ) 2 − 2 s i n 2 θ c o s 2 θ =(1)^{2}-2sin^{2} \theta cos^{2} \theta = ( 1 ) 2 − 2 s i n 2 θ co s 2 θ
= 1 − 2 c o s 2 θ s i n 2 θ =1-2cos^{2} \theta sin^{2} \theta = 1 − 2 co s 2 θ s i n 2 θ
= R H S =RHS = R H S
**प्रश्न 5. Δ A B C \Delta ABC Δ A B C में जिसका कोण B B B समकोण है, A B = 5 c m AB=5 cm A B = 5 c m और ∠ A C B = 30 ∘ \angle ACB=30^{∘} ∠ A C B = 3 0 ∘ (देखिए आकृति)। भुजाओं B C BC B C और A C AC A C की लम्बाइयाँ ज्ञात करें।**
( प्रश्न बैंक )
[Right-angled triangle ABC with angle B=90, angle C=30, AB=5cm]
हल— भुजा B C BC B C की लम्बाई ज्ञात करने के लिए हम उस त्रिकोणमितीय अनुपात को लेंगे जिसमें B C BC B C और दी हुई भुजा A B AB A B हो। क्योंकि B C BC B C कोण C C C की संलग्न भुजा है, और A B AB A B कोण C C C की सम्मुख भुजा है, इसलिए
A B B C = t a n C \frac{AB}{BC}=tanC B C A B = t an C
अर्थात् 5 B C = t a n 30 ∘ = 1 3 \frac{5}{BC}=tan30^{∘}=\frac{1}{\sqrt{3}} B C 5 = t an 3 0 ∘ = 3 1
जिससे B C = 5 3 c m BC=5\sqrt{3} cm B C = 5 3 c m
भुजा A C AC A C
∵ s i n 30 ∘ = A B A C ∵sin30^{∘}=\frac{AB}{AC} ∵ s in 3 0 ∘ = A C A B
अर्थात् 1 2 = 5 A C \frac{1}{2}=\frac{5}{AC} 2 1 = A C 5
अर्थात् A C = 10 c m AC=10 cm A C = 10 c m
**प्रश्न 6. यदि ∠ B \angle B ∠ B और ∠ Q \angle Q ∠ Q ऐसे न्यूनकोण हों जिससे कि s i n B = s i n Q sinB=sinQ s in B = s in Q , तो सिद्ध कीजिए कि ∠ B = ∠ Q \angle B= \angle Q ∠ B = ∠ Q ।**
हल— हम दो समकोण त्रिभुज A B C ABC A B C और P Q R PQR P QR लें, जहाँ s i n B = s i n Q sinB=sinQ s in B = s in Q (देखिए आकृति)।
[Two right-angled triangles ABC and PQR]
यहाँ s i n B = A C A B sinB=\frac{AC}{AB} s in B = A B A C
और s i n Q = P R P Q sinQ=\frac{PR}{PQ} s in Q = P Q P R
तब A C A B = P R P Q \frac{AC}{AB}=\frac{PR}{PQ} A B A C = P Q P R
अतः A C P R = A B P Q = k \frac{AC}{PR}=\frac{AB}{PQ}=k P R A C = P Q A B = k
अब, पाइथागोरस प्रमेय से
B C = A B 2 − A C 2 BC=\sqrt{AB^{2}-AC^{2}} B C = A B 2 − A C 2
और Q R = P Q 2 − P R 2 QR=\sqrt{PQ^{2}-PR^{2}} QR = P Q 2 − P R 2
अतः B C Q R = A B 2 − A C 2 P Q 2 − P R 2 \frac{BC}{QR}=\frac{\sqrt{AB^{2}-AC^{2}}}{\sqrt{PQ^{2}-PR^{2}}} QR B C = P Q 2 − P R 2 A B 2 − A C 2
= k 2 P Q 2 − k 2 P R 2 P Q 2 − P R 2 =\frac{\sqrt{k^{2}PQ^{2}-k^{2}PR^{2}}}{\sqrt{PQ^{2}-PR^{2}}} = P Q 2 − P R 2 k 2 P Q 2 − k 2 P R 2
= k P Q 2 − P R 2 P Q 2 − P R 2 = k =\frac{k\sqrt{PQ^{2}-PR^{2}}}{\sqrt{PQ^{2}-PR^{2}}}=k = P Q 2 − P R 2 k P Q 2 − P R 2 = k
समीकरण (i) और (ii) से
A C P R = A B P Q = B C Q R \frac{AC}{PR}=\frac{AB}{PQ}=\frac{BC}{QR} P R A C = P Q A B = QR B C
तब प्रमेय का प्रयोग करने पर Δ A C B Δ P R Q \Delta ACB~ \Delta PRQ Δ A C B Δ P R Q
अतः ∠ B = ∠ Q \angle B= \angle Q ∠ B = ∠ Q
प्रश्न 7. यदि t a n A = 4 3 tanA=\frac{4}{3} t an A = 3 4 तो निम्नलिखित के मान ज्ञात कीजिए-
(i) s i n A + c o s A sinA+cosA s in A + cos A (ii) c o s 2 A + s i n 2 A cos^{2}A+sin^{2}A co s 2 A + s i n 2 A (iii) c o s 2 A − s i n 2 A cos^{2}A-sin^{2}A co s 2 A − s i n 2 A
[समकोण त्रिभुज ABC जहाँ ∠ B = 90°, AB=3K, BC=4K, AC=5K]
हल— एक समकोण त्रिभुज ABC खींचते हैं।
∴ t a n A = B C A B = 4 3 ∴tanA=\frac{BC}{AB}=\frac{4}{3} ∴ t an A = A B B C = 3 4
अतः यदि B C = 4 K BC=4K B C = 4 K A B = 3 K AB=3K A B = 3 K K K K
पाइथागोरस प्रमेय से
A C 2 = A B 2 + B C 2 AC^{2}=AB^{2}+BC^{2} A C 2 = A B 2 + B C 2
= ( 4 K ) 2 + ( 3 K ) 2 = 25 K 2 =(4K)^{2}+(3K)^{2}=25K^{2} = ( 4 K ) 2 + ( 3 K ) 2 = 25 K 2
∴ A C = 5 K ∴AC=5K ∴ A C = 5 K
∴ ∴ ∴ s i n A = B C A C = 4 5 sinA=\frac{BC}{AC}=\frac{4}{5} s in A = A C B C = 5 4
तथा c o s A = A B A C = 3 5 cosA=\frac{AB}{AC}=\frac{3}{5} cos A = A C A B = 5 3
(ii) c o s 2 A + s i n 2 A = ( 3 5 ) 2 + ( 4 5 ) 2 cos^{2}A+sin^{2}A=(\frac{3}{5})^{2}+(\frac{4}{5})^{2} co s 2 A + s i n 2 A = ( 5 3 ) 2 + ( 5 4 ) 2
= 9 25 + 16 25 = 25 25 = 1 =\frac{9}{25}+\frac{16}{25}=\frac{25}{25}=1 = 25 9 + 25 16 = 25 25 = 1
(iii) c o s 2 A − s i n 2 A = ( 3 5 ) 2 − ( 4 5 ) 2 cos^{2}A-sin^{2}A=(\frac{3}{5})^{2}-(\frac{4}{5})^{2} co s 2 A − s i n 2 A = ( 5 3 ) 2 − ( 5 4 ) 2
= 9 25 − 16 25 = − 7 25 =\frac{9}{25}-\frac{16}{25}=-\frac{7}{25} = 25 9 − 25 16 = − 25 7
प्रश्न 8. निम्न का मान ज्ञात कीजिये-
( 1 + t a n θ + s e c θ ) ( 1 + c o t θ − c o s e c θ ) (1+tan \theta +sec \theta )(1+cot \theta -cosec \theta ) ( 1 + t an θ + sec θ ) ( 1 + co tθ − cosec θ )
हल— ( 1 + t a n θ + s e c θ ) ( 1 + c o t θ − c o s e c θ ) (1+tan \theta +sec \theta )(1+cot \theta -cosec \theta ) ( 1 + t an θ + sec θ ) ( 1 + co tθ − cosec θ )
= ( 1 + s i n θ c o s θ + 1 c o s θ ) ( 1 + c o s θ s i n θ − 1 s i n θ ) =(1+\frac{sin \theta }{cos \theta }+\frac{1}{cos \theta })(1+\frac{cos \theta }{sin \theta }-\frac{1}{sin \theta }) = ( 1 + cos θ s in θ + cos θ 1 ) ( 1 + s in θ cos θ − s in θ 1 )
= ( c o s θ + s i n θ + 1 c o s θ ) ( s i n θ + c o s θ − 1 s i n θ ) =(\frac{cos \theta +sin \theta +1}{cos \theta })(\frac{sin \theta +cos \theta -1}{sin \theta }) = ( cos θ cos θ + s in θ + 1 ) ( s in θ s in θ + cos θ − 1 )
= ( s i n θ + c o s θ ) 2 − ( 1 ) 2 c o s θ s i n θ =\frac{(sin \theta +cos \theta )^{2}-(1)^{2}}{cos \theta sin \theta } = cos θ s in θ ( s in θ + cos θ ) 2 − ( 1 ) 2
= s i n 2 θ + c o s 2 θ + 2 s i n θ c o s θ − 1 c o s θ s i n θ =\frac{sin^{2} \theta +cos^{2} \theta +2sin \theta cos \theta -1}{cos \theta sin \theta } = cos θ s in θ s i n 2 θ + co s 2 θ + 2 s in θ cos θ − 1
= 1 + 2 s i n θ c o s θ − 1 c o s θ s i n θ = 2 s i n θ c o s θ c o s θ s i n θ =\frac{1+2sin \theta cos \theta -1}{cos \theta sin \theta }=\frac{2sin \theta cos \theta }{cos \theta sin \theta } = cos θ s in θ 1 + 2 s in θ cos θ − 1 = cos θ s in θ 2 s in θ cos θ
= 2 =2 = 2
प्रश्न 9. सिद्ध कीजिये-
t a n 2 A − t a n 2 B = c o s 2 B − s i n 2 A c o s 2 B c o s 2 A = s i n 2 A − s i n 2 B c o s 2 A c o s 2 B tan^{2}A-tan^{2}B=\frac{cos^{2}B-sin^{2}A}{cos^{2}Bcos^{2}A}=\frac{sin^{2}A-sin^{2}B}{cos^{2}Acos^{2}B} t a n 2 A − t a n 2 B = co s 2 B co s 2 A co s 2 B − s i n 2 A = co s 2 A co s 2 B s i n 2 A − s i n 2 B
हल— L.H.S. = t a n 2 A − t a n 2 B =tan^{2}A-tan^{2}B = t a n 2 A − t a n 2 B
= s i n 2 A c o s 2 A − s i n 2 B c o s 2 B =\frac{sin^{2}A}{cos^{2}A}-\frac{sin^{2}B}{cos^{2}B} = co s 2 A s i n 2 A − co s 2 B s i n 2 B
= s i n 2 A c o s 2 B − c o s 2 A s i n 2 B c o s 2 A c o s 2 B =\frac{sin^{2}Acos^{2}B-cos^{2}Asin^{2}B}{cos^{2}Acos^{2}B} = co s 2 A co s 2 B s i n 2 A co s 2 B − co s 2 A s i n 2 B
= ( 1 − c o s 2 A ) c o s 2 B − c o s 2 A ( 1 − c o s 2 B ) c o s 2 A c o s 2 B =\frac{(1-cos^{2}A)cos^{2}B-cos^{2}A(1-cos^{2}B)}{cos^{2}Acos^{2}B} = co s 2 A co s 2 B ( 1 − co s 2 A ) co s 2 B − co s 2 A ( 1 − co s 2 B )
= c o s 2 B − c o s 2 A c o s 2 B − c o s 2 A + c o s 2 A c o s 2 B c o s 2 A c o s 2 B =\frac{cos^{2}B-cos^{2}Acos^{2}B-cos^{2}A+cos^{2}Acos^{2}B}{cos^{2}Acos^{2}B} = co s 2 A co s 2 B co s 2 B − co s 2 A co s 2 B − co s 2 A + co s 2 A co s 2 B
= c o s 2 B − c o s 2 A c o s 2 A c o s 2 B =\frac{cos^{2}B-cos^{2}A}{cos^{2}Acos^{2}B} = co s 2 A co s 2 B co s 2 B − co s 2 A
= ( 1 − s i n 2 B ) − ( 1 − s i n 2 A ) c o s 2 A c o s 2 B =\frac{(1-sin^{2}B)-(1-sin^{2}A)}{cos^{2}Acos^{2}B} = co s 2 A co s 2 B ( 1 − s i n 2 B ) − ( 1 − s i n 2 A )
= s i n 2 A − s i n 2 B c o s 2 A c o s 2 B = R . H . S . =\frac{sin^{2}A-sin^{2}B}{cos^{2}Acos^{2}B}=R.H.S. = co s 2 A co s 2 B s i n 2 A − s i n 2 B = R . H . S .
प्रश्न 10. यदि t a n θ = 1 7 tan \theta =\frac{1}{\sqrt{7}} t an θ = 7 1 , तो c o s e c 2 θ − s e c 2 θ c o s e c 2 θ + s e c 2 θ \frac{cosec^{2} \theta -sec^{2} \theta }{cosec^{2} \theta +sec^{2} \theta } cose c 2 θ + se c 2 θ cose c 2 θ − se c 2 θ का मान लिखिए।
हल— यहाँ t a n θ = 1 7 tan \theta =\frac{1}{\sqrt{7}} t an θ = 7 1
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हम जानते हैं कि s e c 2 θ = 1 + t a n 2 θ sec^{2} \theta =1+tan^{2} \theta se c 2 θ = 1 + t a n 2 θ ∴ ∴ ∴
= 1 + 1 7 2 =1+\frac{1}{\sqrt{7}}^{2} = 1 + 7 1 2 प्रश्न 12. सर्वसमिका s e c 2 θ = 1 + t a n 2 θ sec^{2} \theta =1+tan^{2} \theta se c 2 θ = 1 + t a n 2 θ
= 1 + 1 7 = 8 7 =1+\frac{1}{7}=\frac{8}{7} = 1 + 7 1 = 7 8 s i n θ − c o s θ + 1 s i n θ + c o s θ − 1 = 1 s e c θ − t a n θ \frac{sin \theta -cos \theta +1}{sin \theta +cos \theta -1}=\frac{1}{sec \theta -tan \theta } s in θ + cos θ − 1 s in θ − cos θ + 1 = sec θ − t an θ 1
पुन: c o s e c 2 θ = 1 + c o t 2 θ cosec^{2} \theta =1+cot^{2} \theta cose c 2 θ = 1 + co t 2 θ हल —क्योंकि हमें s e c θ sec \theta sec θ t a n θ tan \theta t an θ c o s θ cos \theta cos θ s e c θ sec \theta sec θ t a n θ tan \theta t an θ
∵ t a n θ = 1 c o t θ = 1 7 ∵tan \theta =\frac{1}{cot \theta }=\frac{1}{\sqrt{7}} ∵ t an θ = co tθ 1 = 7 1 = s i n θ − c o s θ + 1 s i n θ + c o s θ − 1 = t a n θ − 1 + s e c θ t a n θ + 1 − s e c θ =\frac{sin \theta -cos \theta +1}{sin \theta +cos \theta -1}=\frac{tan \theta -1+sec \theta }{tan \theta +1-sec \theta } = s in θ + cos θ − 1 s in θ − cos θ + 1 = t an θ + 1 − sec θ t an θ − 1 + sec θ
अतः c o t θ = ( 7 ) cot \theta =(\sqrt{7}) co tθ = ( 7 ) = ( t a n θ + s e c θ ) − 1 ( t a n θ − s e c θ ) + 1 =\frac{(tan \theta +sec \theta )-1}{(tan \theta -sec \theta )+1} = ( t an θ − sec θ ) + 1 ( t an θ + sec θ ) − 1
= 1 + ( 7 ) 2 =1+(\sqrt{7})^{2} = 1 + ( 7 ) 2 = ( t a n θ + s e c θ ) − 1 ( t a n θ − s e c θ ) ( t a n θ − s e c θ ) + 1 ( t a n θ − s e c θ ) =\frac{{(tan \theta +sec \theta )-1}(tan \theta -sec \theta )}{{(tan \theta -sec \theta )+1}(tan \theta -sec \theta )} = ( t an θ − sec θ ) + 1 ( t an θ − sec θ ) ( t an θ + sec θ ) − 1 ( t an θ − sec θ )
= 1 + 7 = 8 =1+7=8 = 1 + 7 = 8 ( t a n θ ) (tan \theta ) ( t an θ )
व्यंजक में मान रखने पर c o s e c 2 θ − s e c 2 θ c o s e c 2 θ + s e c 2 θ \frac{cosec^{2} \theta -sec^{2} \theta }{cosec^{2} \theta +sec^{2} \theta } cose c 2 θ + se c 2 θ cose c 2 θ − se c 2 θ = ( t a n 2 θ − s e c 2 θ ) − ( t a n θ − s e c θ ) t a n θ − s e c θ + 1 ( t a n θ − s e c θ ) =\frac{(tan^{2} \theta -sec^{2} \theta )-(tan \theta -sec \theta )}{{tan \theta -sec \theta +1}(tan \theta -sec \theta )} = t an θ − sec θ + 1 ( t an θ − sec θ ) ( t a n 2 θ − se c 2 θ ) − ( t an θ − sec θ )
= 8 − 8 7 8 + 8 7 = 48 64 = 3 4 =\frac{8-\frac{8}{7}}{8+\frac{8}{7}}=\frac{48}{64}=\frac{3}{4} = 8 + 7 8 8 − 7 8 = 64 48 = 4 3 = − 1 − t a n θ + s e c θ ( t a n θ − s e c θ + 1 ) ( t a n θ − s e c θ ) =\frac{-1-tan \theta +sec \theta }{(tan \theta -sec \theta +1)(tan \theta -sec \theta )} = ( t an θ − sec θ + 1 ) ( t an θ − sec θ ) − 1 − t an θ + sec θ
प्रश्न 11. निम्न सर्वसमिका को सिद्ध कीजिये— | ∵ s e c 2 θ − t a n 2 θ = 1 ∵sec^{2} \theta -tan^{2} \theta =1 ∵ se c 2 θ − t a n 2 θ = 1
s i n A + c o s A s i n A − c o s A + s i n A − c o s A s i n A + c o s A = 2 s i n 2 A − c o s 2 A \frac{sinA+cosA}{sinA-cosA}+\frac{sinA-cosA}{sinA+cosA}=\frac{2}{sin^{2}A-cos^{2}A} s in A − cos A s in A + cos A + s in A + cos A s in A − cos A = s i n 2 A − co s 2 A 2 = − ( 1 + t a n θ − s e c θ ) ( t a n θ − s e c θ + 1 ) ( t a n θ − s e c θ ) =\frac{-(1+tan \theta -sec \theta )}{(tan \theta -sec \theta +1)(tan \theta -sec \theta )} = ( t an θ − sec θ + 1 ) ( t an θ − sec θ ) − ( 1 + t an θ − sec θ )
हल— L.H.S. = s i n A + c o s A s i n A − c o s A + s i n A − c o s A s i n A + c o s A =\frac{sinA+cosA}{sinA-cosA}+\frac{sinA-cosA}{sinA+cosA} = s in A − cos A s in A + cos A + s in A + cos A s in A − cos A | = − 1 t a n θ − s e c θ = 1 s e c θ − t a n θ =\frac{-1}{tan \theta -sec \theta }=\frac{1}{sec \theta -tan \theta } = t an θ − sec θ − 1 = sec θ − t an θ 1 (इतिसिद्धम्)
= ( s i n A + c o s A ) 2 + ( s i n A − c o s A ) 2 ( s i n A − c o s A ) ( s i n A + c o s A ) =\frac{(sinA+cosA)^{2}+(sinA-cosA)^{2}}{(sinA-cosA)(sinA+cosA)} = ( s in A − cos A ) ( s in A + cos A ) ( s in A + cos A ) 2 + ( s in A − cos A ) 2 प्रश्न 13. सिद्ध कीजिए कि (माध्य. शिक्षा बोर्ड, मॉडल पेपर, 2024-25)
= s i n 2 A + 2 s i n A c o s A + c o s 2 A + s i n 2 A − 2 s i n A c o s A + c o s 2 A s i n 2 A − c o s 2 A =\frac{sin^{2}A+2sinAcosA+cos^{2}A+sin^{2}A-2sinAcosA+cos^{2}A}{sin^{2}A-cos^{2}A} = s i n 2 A − co s 2 A s i n 2 A + 2 s in A cos A + co s 2 A + s i n 2 A − 2 s in A cos A + co s 2 A 1 + c o s A 1 − c o s A = c o s e c A + c o t A \sqrt{\frac{1+cosA}{1-cosA}}=cosec A+cotA 1 − cos A 1 + cos A = cosec A + co t A
= 2 s i n 2 A + 2 c o s 2 A s i n 2 A − c o s 2 A = 2 ( s i n 2 A + c o s 2 A ) s i n 2 A − c o s 2 A =\frac{2sin^{2}A+2cos^{2}A}{sin^{2}A-cos^{2}A}=\frac{2(sin^{2}A+cos^{2}A)}{sin^{2}A-cos^{2}A} = s i n 2 A − co s 2 A 2 s i n 2 A + 2 co s 2 A = s i n 2 A − co s 2 A 2 ( s i n 2 A + co s 2 A ) हल —L.H.S. = 1 + c o s A 1 − c o s A =\sqrt{\frac{1+cosA}{1-cosA}} = 1 − cos A 1 + cos A
= 2 × 1 s i n 2 A − c o s 2 A =\frac{2 \times 1}{sin^{2}A-cos^{2}A} = s i n 2 A − co s 2 A 2 × 1 ∵ s i n 2 A + c o s 2 A = 1 ∵sin^{2}A+cos^{2}A=1 ∵ s i n 2 A + co s 2 A = 1 1 + c o s A 1+cosA 1 + cos A
= 2 s i n 2 A − c o s 2 A = R . H . S . =\frac{2}{sin^{2}A-cos^{2}A}=R.H.S. = s i n 2 A − co s 2 A 2 = R . H . S . = ( 1 + c o s A ) ( 1 + c o s A ) ( 1 − c o s A ) ( 1 + c o s A ) =\sqrt{\frac{(1+cosA)(1+cosA)}{(1-cosA)(1+cosA)}} = ( 1 − cos A ) ( 1 + cos A ) ( 1 + cos A ) ( 1 + cos A )
| = ( 1 + c o s A ) 2 1 − c o s 2 A = ( 1 + c o s A ) 2 s i n 2 A =\sqrt{\frac{(1+cosA)^{2}}{1-cos^{2}A}}=\sqrt{\frac{(1+cosA)^{2}}{sin^{2}A}} = 1 − co s 2 A ( 1 + cos A ) 2 = s i n 2 A ( 1 + cos A ) 2
( ∵ s i n 2 A = 1 − c o s 2 A ) (∵sin^{2}A=1-cos^{2}A) ( ∵ s i n 2 A = 1 − co s 2 A )
= 1 + c o s A s i n A 2 = 1 + c o s A s i n A =\sqrt{\frac{1+cosA}{sinA}^{2}}=\frac{1+cosA}{sinA} = s in A 1 + cos A 2 = s in A 1 + cos A
= 1 s i n A + c o s A s i n A =\frac{1}{sinA}+\frac{cosA}{sinA} = s in A 1 + s in A cos A
= c o s e c A + c o t A =cosec A+cotA = cosec A + co t A
= R . H . S . =R.H.S. = R . H . S .
∴ L . H . S . = R . H . S . ∴L.H.S.=R.H.S. ∴ L . H . S . = R . H . S .
प्रश्न 14. निम्नलिखित सर्वसमिका सिद्ध कीजिए, जहाँ वे कोण, जिनके लिए व्यंजक परिभाषित है, न्यून कोण है।
1 + c o t 2 A 1 + t a n 2 A = 1 − c o t A 1 − t a n A 2 \frac{1+cot^{2}A}{1+tan^{2}A}=\frac{1-cotA}{1-tanA}^{2} 1 + t a n 2 A 1 + co t 2 A = 1 − t an A 1 − co t A 2
हल—
L . H . S . = 1 + c o t 2 A 1 + t a n 2 A L.H.S.=\frac{1+cot^{2}A}{1+tan^{2}A} L . H . S . = 1 + t a n 2 A 1 + co t 2 A
= c o s e c 2 A s e c 2 A =\frac{cosec^{2}A}{sec^{2}A} = se c 2 A cose c 2 A
= 1 s i n 2 A 1 c o s 2 A = c o s 2 A s i n 2 A = c o t 2 A =\frac{\frac{1}{sin^{2}A}}{\frac{1}{cos^{2}A}}=\frac{cos^{2}A}{sin^{2}A}=cot^{2}A = co s 2 A 1 s i n 2 A 1 = s i n 2 A co s 2 A = co t 2 A
R . H . S . = 1 − c o t A 1 − t a n A 2 = 1 − c o s A s i n A 1 − s i n A c o s A 2 R.H.S.=\frac{1-cotA}{1-tanA}^{2}=\frac{1-\frac{cosA}{sinA}}{1-\frac{sinA}{cosA}}^{2} R . H . S . = 1 − t an A 1 − co t A 2 = 1 − cos A s in A 1 − s in A cos A 2
= s i n A − c o s A s i n A c o s A − s i n A c o s A 2 =\frac{\frac{sinA-cosA}{sinA}}{\frac{cosA-sinA}{cosA}}^{2} = cos A cos A − s in A s in A s in A − cos A 2
= ( s i n A − c o s A ) × c o s A − ( s i n A − c o s A ) × s i n A 2 =\frac{(sinA-cosA) \times cosA}{-(sinA-cosA) \times sinA}^{2} = − ( s in A − cos A ) × s in A ( s in A − cos A ) × cos A 2
= c o s A − s i n A 2 = c o s 2 A s i n 2 A = c o t 2 A =\frac{cosA}{-sinA}^{2}=\frac{cos^{2}A}{sin^{2}A}=cot^{2}A = − s in A cos A 2 = s i n 2 A co s 2 A = co t 2 A
∴ L . H . S . = R . H . S . ∴L.H.S.=R.H.S. ∴ L . H . S . = R . H . S .
प्रश्न 15. यदि s i n θ + c o s θ = p sin \theta +cos \theta =p s in θ + cos θ = p और s e c θ + c o s e c θ = q sec \theta +cosec \theta =q sec θ + cosec θ = q हो, तो सिद्ध कीजिए कि q ( p 2 − 1 ) = 2 p q(p^{2}-1)=2p q ( p 2 − 1 ) = 2 p
हल— L . H . S . = q ( p 2 − 1 ) L.H.S.=q(p^{2}-1) L . H . S . = q ( p 2 − 1 )
p p p q q q
= ( s e c θ + c o s e c θ ) ( ( s i n θ + c o s θ ) 2 − 1 ) =(sec \theta +cosec \theta )((sin \theta +cos \theta )^{2}-1) = ( sec θ + cosec θ ) (( s in θ + cos θ ) 2 − 1 )
= 1 c o s θ + 1 s i n θ [ s i n 2 θ + c o s 2 θ + 2 s i n θ c o s θ − 1 ] =\frac{1}{cos \theta }+\frac{1}{sin \theta }[sin^{2} \theta +cos^{2} \theta +2sin \theta cos \theta -1] = cos θ 1 + s in θ 1 [ s i n 2 θ + co s 2 θ + 2 s in θ cos θ − 1 ]
= s i n θ + c o s θ c o s θ s i n θ [ 1 + 2 s i n θ c o s θ − 1 ] =\frac{sin \theta +cos \theta }{cos \theta sin \theta }[1+2sin \theta cos \theta -1] = cos θ s in θ s in θ + cos θ [ 1 + 2 s in θ cos θ − 1 ]
= s i n θ + c o s θ c o s θ s i n θ × [ 2 s i n θ c o s θ ] =\frac{sin \theta +cos \theta }{cos \theta sin \theta } \times [2sin \theta cos \theta ] = cos θ s in θ s in θ + cos θ × [ 2 s in θ cos θ ]
= 2 [ s i n θ + c o s θ ] = 2 p = L . H . S . =2[sin \theta +cos \theta ]=2p=L.H.S. = 2 [ s in θ + cos θ ] = 2 p = L . H . S .
प्रश्न 16. सिद्ध कीजिए कि
c o t A + c o s e c A − 1 c o t A − c o s e c A + 1 = 1 + c o s A s i n A \frac{cotA+cosec A-1}{cotA-cosec A+1}=\frac{1+cosA}{sinA} co t A − cosec A + 1 co t A + cosec A − 1 = s in A 1 + cos A
हल— L . H . S . = c o t A + c o s e c A − 1 c o t A − c o s e c A + 1 L.H.S.=\frac{cotA+cosec A-1}{cotA-cosec A+1} L . H . S . = co t A − cosec A + 1 co t A + cosec A − 1
= ( c o t A + c o s e c A ) − ( c o s e c 2 A − c o t 2 A ) c o t A − c o s e c A + 1 =\frac{(cotA+cosec A)-(cosec^{2}A-cot^{2}A)}{cotA-cosec A+1} = co t A − cosec A + 1 ( co t A + cosec A ) − ( cose c 2 A − co t 2 A ) ( ∵ c o s e c 2 A − c o t 2 A = 1 ) (∵cosec^{2}A-cot^{2}A=1) ( ∵ cose c 2 A − co t 2 A = 1 )
= ( c o s e c A + c o t A ) − [ ( c o s e c A + c o t A ) ( c o s e c A − c o t A ) ] c o t A − c o s e c A + 1 =\frac{(cosec A+cotA)-[(cosec A+cotA)(cosec A-cotA)]}{cotA-cosec A+1} = co t A − cosec A + 1 ( cosec A + co t A ) − [( cosec A + co t A ) ( cosec A − co t A )]
= ( c o s e c A + c o t A ) [ 1 − ( c o s e c A − c o t A ) ] c o t A − c o s e c A + 1 =\frac{(cosec A+cotA)[1-(cosec A-cotA)]}{cotA-cosec A+1} = co t A − cosec A + 1 ( cosec A + co t A ) [ 1 − ( cosec A − co t A )]
= ( c o s e c A + c o t A ) [ 1 − c o s e c A + c o t A ] ( c o t A − c o s e c A + 1 ) =\frac{(cosec A+cotA)[1-cosec A+cotA]}{(cotA-cosec A+1)} = ( co t A − cosec A + 1 ) ( cosec A + co t A ) [ 1 − cosec A + co t A ]
= c o s e c A + c o t A =cosec A+cotA = cosec A + co t A
= 1 s i n A + c o s A s i n A = 1 + c o s A s i n A = R . H . S . =\frac{1}{sinA}+\frac{cosA}{sinA}=\frac{1+cosA}{sinA}=R.H.S. = s in A 1 + s in A cos A = s in A 1 + cos A = R . H . S .
**प्रश्न 17. यदि s e c θ + t a n θ = p sec \theta +tan \theta =p sec θ + t an θ = p हो, तो सिद्ध करो कि p 2 − 1 p 2 + 1 = s i n θ \frac{p^{2}-1}{p^{2}+1}=sin \theta p 2 + 1 p 2 − 1 = s in θ
हल— L . H . S . = p 2 − 1 p 2 + 1 = ( s e c θ + t a n θ ) 2 − 1 ) ( s e c θ ) L.H.S.=\frac{p^{2}-1}{p^{2}+1}=\frac{(sec \theta +tan \theta )^{2}-1)}{(sec \theta )} L . H . S . = p 2 + 1 p 2 − 1 = ( sec θ ) ( sec θ + t an θ ) 2 − 1 )
= s e c 2 θ + 2 s e c θ t a n θ + t a n 2 θ − 1 s e c 2 θ + 2 s e c θ t a n θ + t a n 2 θ + 1 =\frac{sec^{2} \theta +2sec \theta tan \theta +tan^{2} \theta -1}{sec^{2} \theta +2sec \theta tan \theta +tan^{2} \theta +1} = se c 2 θ + 2 sec θ t an θ + t a n 2 θ + 1 se c 2 θ + 2 sec θ t an θ + t a n 2 θ − 1
( ∵ s e c 2 θ − 1 = t a n 2 θ ) (∵sec^{2} \theta -1=tan^{2} \theta ) ( ∵ se c 2 θ − 1 = t a n 2 θ )
तथा 1 + t a n 2 θ = s e c 2 θ 1+tan^{2} \theta =sec^{2} \theta 1 + t a n 2 θ = se c 2 θ
= p 2 − 1 p 2 + 1 = t a n 2 θ + 2 s e c θ t a n θ + t a n 2 θ s e c 2 θ + 2 s e c θ t a n θ + s e c 2 θ =\frac{p^{2}-1}{p^{2}+1}=\frac{tan^{2} \theta +2sec \theta tan \theta +tan^{2} \theta }{sec^{2} \theta +2sec \theta tan \theta +sec^{2} \theta } = p 2 + 1 p 2 − 1 = se c 2 θ + 2 sec θ t an θ + se c 2 θ t a n 2 θ + 2 sec θ t an θ + t a n 2 θ
= 2 t a n 2 θ + 2 s e c θ t a n θ 2 s e c 2 θ + 2 s e c θ t a n θ =\frac{2tan^{2} \theta +2sec \theta tan \theta }{2sec^{2} \theta +2sec \theta tan \theta } = 2 se c 2 θ + 2 sec θ t an θ 2 t a n 2 θ + 2 sec θ t an θ
= 2 t a n θ [ t a n θ + s e c θ ] 2 s e c θ [ s e c θ + t a n θ ] =\frac{2tan \theta [tan \theta +sec \theta ]}{2sec \theta [sec \theta +tan \theta ]} = 2 sec θ [ sec θ + t an θ ] 2 t an θ [ t an θ + sec θ ]
= t a n θ s e c θ = s i n θ c o s θ 1 c o s θ =\frac{tan \theta }{sec \theta }=\frac{\frac{sin \theta }{cos \theta }}{\frac{1}{cos \theta }} = sec θ t an θ = cos θ 1 cos θ s in θ
= s i n θ ⋅ c o s θ c o s θ = s i n θ = R . H . S . =\frac{sin \theta ·cos \theta }{cos \theta }=sin \theta =R.H.S. = cos θ s in θ ⋅ cos θ = s in θ = R . H . S .
प्रश्न 18. यदि c o s A c o s B = m \frac{cosA}{cosB}=m cos B cos A = m तथा c o s A s i n B = n \frac{cosA}{sinB}=n s in B cos A = n हो, तो सिद्ध कीजिये ( m 2 + n 2 ) c o s 2 B = n 2 (m^{2}+n^{2})cos^{2}B=n^{2} ( m 2 + n 2 ) co s 2 B = n 2
हल— L.H.S. ( m 2 + n 2 ) c o s 2 B (m^{2}+n^{2})cos^{2}B ( m 2 + n 2 ) co s 2 B मान रखने पर
⇒ c o s A c o s B 2 + c o s A s i n B 2 c o s 2 B ⇒\frac{cosA}{cosB}^{2}+\frac{cosA}{sinB}^{2}cos^{2}B ⇒ cos B cos A 2 + s in B cos A 2 co s 2 B
⇒ c o s 2 A s i n 2 B + c o s 2 B c o s 2 A c o s 2 B s i n 2 B c o s 2 B ⇒\frac{cos^{2}Asin^{2}B+cos^{2}Bcos^{2}A}{cos^{2}Bsin^{2}B}cos^{2}B ⇒ co s 2 B s i n 2 B co s 2 A s i n 2 B + co s 2 B co s 2 A co s 2 B
⇒ c o s 2 A ( 1 − c o s 2 B ) + c o s 2 B c o s 2 A s i n 2 B ⇒\frac{cos^{2}A(1-cos^{2}B)+cos^{2}Bcos^{2}A}{sin^{2}B} ⇒ s i n 2 B co s 2 A ( 1 − co s 2 B ) + co s 2 B co s 2 A
⇒ c o s 2 A − c o s 2 A c o s 2 B + c o s 2 B c o s 2 A s i n 2 B ⇒\frac{cos^{2}A-cos^{2}Acos^{2}B+cos^{2}Bcos^{2}A}{sin^{2}B} ⇒ s i n 2 B co s 2 A − co s 2 A co s 2 B + co s 2 B co s 2 A
⇒ c o s 2 A s i n 2 B = c o s A s i n B 2 = n 2 = R . H . S . ⇒\frac{cos^{2}A}{sin^{2}B}=\frac{cosA}{sinB}^{2}=n^{2}=R.H.S. ⇒ s i n 2 B co s 2 A = s in B cos A 2 = n 2 = R . H . S .
प्रश्न 19. सिद्ध कीजिये कि 1 − c o s A 1 + c o s A = c o s e c A − c o t A \sqrt{\frac{1-cosA}{1+cosA}}=cosec A-cotA 1 + cos A 1 − cos A = cosec A − co t A ( प्रश्न बैंक; माध्य. शिक्षा बोर्ड, 2023 )
हल— L.H.S. = 1 − c o s A 1 + c o s A =\sqrt{\frac{1-cosA}{1+cosA}} = 1 + cos A 1 − cos A
वर्गमूल के अन्दर अंश व हर में ( 1 ) (1) ( 1 )
= ( 1 − c o s A ) ( 1 − c o s A ) ( 1 + c o s A ) ( 1 − c o s A ) = ( 1 − c o s A ) 2 1 − c o s 2 A =\sqrt{\frac{(1-cosA)(1-cosA)}{(1+cosA)(1-cosA)}}=\sqrt{\frac{(1-cosA)^{2}}{1-cos^{2}A}} = ( 1 + cos A ) ( 1 − cos A ) ( 1 − cos A ) ( 1 − cos A ) = 1 − co s 2 A ( 1 − cos A ) 2
= ( 1 − c o s A ) 2 s i n 2 A =\sqrt{\frac{(1-cosA)^{2}}{sin^{2}A}} = s i n 2 A ( 1 − cos A ) 2 ∵ s i n 2 A = 1 − c o s 2 A ∵sin^{2}A=1-cos^{2}A ∵ s i n 2 A = 1 − co s 2 A
= 1 − c o s A s i n A = 1 s i n A − c o s A s i n A =\frac{1-cosA}{sinA}=\frac{1}{sinA}-\frac{cosA}{sinA} = s in A 1 − cos A = s in A 1 − s in A cos A
= c o s e c A − c o t A = R . H . S . =cosec A-cotA=R.H.S. = cosec A − co t A = R . H . S .
∴ L . H . S . = R . H . S . ∴L.H.S.=R.H.S. ∴ L . H . S . = R . H . S .
प्रश्न 20. सिद्ध कीजिये कि ( c o s e c A − s i n A ) ( s e c A − c o s A ) ( t a n A + c o t A ) = 1 (cosec A-sinA)(secA-cosA)(tanA+cotA)=1 ( cosec A − s in A ) ( sec A − cos A ) ( t an A + co t A ) = 1 . ( माध्य. शिक्षा बोर्ड, 2024 )
हल— L.H.S. ( c o s e c A − s i n A ) ( s e c A − c o s A ) ( t a n A + c o t A ) (cosec A-sinA)(secA-cosA)(tanA+cotA) ( cosec A − s in A ) ( sec A − cos A ) ( t an A + co t A )
= 1 s i n A − s i n A 1 c o s A − c o s A s i n A c o s A + c o s A s i n A =\frac{1}{sinA}-sinA\frac{1}{cosA}-cosA\frac{sinA}{cosA}+\frac{cosA}{sinA} = s in A 1 − s in A cos A 1 − cos A cos A s in A + s in A cos A
= ( 1 − s i n 2 A ) ( 1 − c o s 2 A ) s i n A c o s A s i n 2 A + c o s 2 A s i n A c o s A =\frac{(1-sin^{2}A)(1-cos^{2}A)}{sinAcosA}\frac{sin^{2}A+cos^{2}A}{sinAcosA} = s in A cos A ( 1 − s i n 2 A ) ( 1 − co s 2 A ) s in A cos A s i n 2 A + co s 2 A
= c o s 2 A s i n 2 A s i n A c o s A × 1 s i n A c o s A =\frac{cos^{2}Asin^{2}A}{sinAcosA} \times \frac{1}{sinAcosA} = s in A cos A co s 2 A s i n 2 A × s in A cos A 1
∵ sin^2 A + cos^2 A = 1
∵ 1 - sin^2 A = cos^2 A
1 - cos^2 A = sin^2 A
= c o s 2 A s i n 2 A s i n 2 A c o s 2 A = 1 = R . H . S . =\frac{cos^{2}Asin^{2}A}{sin^{2}Acos^{2}A}=1=R.H.S. = s i n 2 A co s 2 A co s 2 A s i n 2 A = 1 = R . H . S .
अतः L.H.S. = R.H.S. ( इतिसिद्धम् )
प्रश्न 21. सिद्ध कीजिये कि s i n θ 1 + c o s θ + 1 + c o s θ s i n θ = 2 c o s e c θ \frac{sin \theta }{1+cos \theta }+\frac{1+cos \theta }{sin \theta }=2cosec \theta 1 + cos θ s in θ + s in θ 1 + cos θ = 2 cosec θ ( प्रश्न बैंक )
हल— L.H.S. = s i n θ 1 + c o s θ + 1 + c o s θ s i n θ =\frac{sin \theta }{1+cos \theta }+\frac{1+cos \theta }{sin \theta } = 1 + cos θ s in θ + s in θ 1 + cos θ
= s i n 2 θ + ( 1 + c o s θ ) 2 ( 1 + c o s θ ) s i n θ =\frac{sin^{2} \theta +(1+cos \theta )^{2}}{(1+cos \theta )sin \theta } = ( 1 + cos θ ) s in θ s i n 2 θ + ( 1 + cos θ ) 2
= s i n 2 θ + 1 + c o s 2 θ + 2 c o s θ ( 1 + c o s θ ) s i n θ =\frac{sin^{2} \theta +1+cos^{2} \theta +2cos \theta }{(1+cos \theta )sin \theta } = ( 1 + cos θ ) s in θ s i n 2 θ + 1 + co s 2 θ + 2 cos θ
= 1 + 1 + 2 c o s θ ( 1 + c o s θ ) s i n θ = 2 + 2 c o s θ ( 1 + c o s θ ) s i n θ =\frac{1+1+2cos \theta }{(1+cos \theta )sin \theta }=\frac{2+2cos \theta }{(1+cos \theta )sin \theta } = ( 1 + cos θ ) s in θ 1 + 1 + 2 cos θ = ( 1 + cos θ ) s in θ 2 + 2 cos θ
= 2 ( 1 + c o s θ ) ( 1 + c o s θ ) s i n θ = 2 s i n θ =\frac{2(1+cos \theta )}{(1+cos \theta )sin \theta }=\frac{2}{sin \theta } = ( 1 + cos θ ) s in θ 2 ( 1 + cos θ ) = s in θ 2
= 2 c s c θ = R . H . S . =2csc \theta =R.H.S. = 2 csc θ = R . H . S .
प्रश्न 22. सिद्ध कीजिए कि— (प्रश्न बैंक)
1 − s i n θ 1 + s i n θ = s e c θ − t a n θ \sqrt{\frac{1-sin \theta }{1+sin \theta }}=sec \theta -tan \theta 1 + s in θ 1 − s in θ = sec θ − t an θ
हल— L . H . S . = 1 − s i n θ 1 + s i n θ L.H.S.=\sqrt{\frac{1-sin \theta }{1+sin \theta }} L . H . S . = 1 + s in θ 1 − s in θ
= ( 1 ) ( 1 ) × ( 1 ) ( 1 ) =\sqrt{\frac{(1)}{(1)} \times \frac{(1)}{(1)}} = ( 1 ) ( 1 ) × ( 1 ) ( 1 )
= ( 1 − s i n θ ) 2 1 − s i n 2 θ =\sqrt{\frac{(1-sin \theta )^{2}}{1-sin^{2} \theta }} = 1 − s i n 2 θ ( 1 − s in θ ) 2
= 1 − s i n θ c o s θ = 1 c o s θ − s i n θ c o s θ =\frac{1-sin \theta }{cos \theta }=\frac{1}{cos \theta }-\frac{sin \theta }{cos \theta } = cos θ 1 − s in θ = cos θ 1 − cos θ s in θ
= s e c θ − t a n θ = R . H . S . =sec \theta -tan \theta =R.H.S. = sec θ − t an θ = R . H . S .
प्रश्न 23. यदि c o s θ = 3 5 cos \theta =\frac{3}{5} cos θ = 5 3 , तो s i n θ t a n θ − 1 2 t a n 2 θ \frac{sin \theta tan \theta -1}{2tan^{2} \theta } 2 t a n 2 θ s in θ t an θ − 1 का मान ज्ञात कीजिए। (प्रश्न बैंक)
हल— दिया है, c o s θ = 3 5 cos \theta =\frac{3}{5} cos θ = 5 3
∴ s i n θ = 4 5 , t a n θ = 4 3 ∴sin \theta =\frac{4}{5},tan \theta =\frac{4}{3} ∴ s in θ = 5 4 , t an θ = 3 4
∴ s i n θ t a n θ − 1 2 t a n 2 θ = 4 5 × 4 3 − 1 2 × 4 3 2 ∴\frac{sin \theta tan \theta -1}{2tan^{2} \theta }=\frac{\frac{4}{5} \times \frac{4}{3}-1}{2 \times \frac{4}{3}^{2}} ∴ 2 t a n 2 θ s in θ t an θ − 1 = 2 × 3 4 2 5 4 × 3 4 − 1
= 16 − 15 15 2 × 16 9 = 1 15 32 9 = 1 15 × 9 32 = 3 160 =\frac{\frac{16-15}{15}}{2 \times \frac{16}{9}}=\frac{\frac{1}{15}}{\frac{32}{9}}=\frac{1}{15} \times \frac{9}{32}=\frac{3}{160} = 2 × 9 16 15 16 − 15 = 9 32 15 1 = 15 1 × 32 9 = 160 3
प्रश्न 24. त्रिभुज A C B ACB A C B जिसका कोण C C C समकोण है जिसमें A B = 29 AB=29 A B = 29 इकाई, B C = 21 BC=21 B C = 21 इकाई और ∠ A B C = θ \angle ABC= \theta ∠ A B C = θ है तो निम्नांकित के मान ज्ञात कीजिए— (प्रश्न बैंक)
(i) c o s 2 θ + s i n 2 θ cos^{2} \theta +sin^{2} \theta co s 2 θ + s i n 2 θ
(ii) c o s 2 θ − s i n 2 θ cos^{2} \theta -sin^{2} \theta co s 2 θ − s i n 2 θ
हल— समकोण Δ A B C \Delta ABC Δ A B C में, A B 2 = A C 2 + B C 2 AB^{2}=AC^{2}+BC^{2} A B 2 = A C 2 + B C 2
∴ A C 2 = ( 29 ) 2 − ( 21 ) 2 ∴AC^{2}=(29)^{2}-(21)^{2} ∴ A C 2 = ( 29 ) 2 − ( 21 ) 2
= ( 29 − 21 ) ( 29 + 21 ) =(29-21)(29+21) = ( 29 − 21 ) ( 29 + 21 )
A C = 8 × 50 AC=\sqrt{8 \times 50} A C = 8 × 50
A C = 20 AC=20 A C = 20
∴ s i n θ = 20 29 ∴sin \theta =\frac{20}{29} ∴ s in θ = 29 20 c o s θ = 21 29 cos \theta =\frac{21}{29} cos θ = 29 21
[Right angled triangle ABC with C at right angle, AC=20, BC=21, AB=29, angle B = theta]
(i) c o s 2 θ + s i n 2 θ = 21 29 2 + 20 29 2 cos^{2} \theta +sin^{2} \theta =\frac{21}{29}^{2}+\frac{20}{29}^{2} co s 2 θ + s i n 2 θ = 29 21 2 + 29 20 2
= 441 + 400 841 = 841 841 = 1 =\frac{441+400}{841}=\frac{841}{841}=1 = 841 441 + 400 = 841 841 = 1
(ii) c o s 2 θ − s i n 2 θ = 21 29 2 − 20 29 2 cos^{2} \theta -sin^{2} \theta =\frac{21}{29}^{2}-\frac{20}{29}^{2} co s 2 θ − s i n 2 θ = 29 21 2 − 29 20 2
= ( 21 + 20 ) ( 21 − 20 ) ( 29 ) 2 = 41 841 =\frac{(21+20)(21-20)}{(29)^{2}}=\frac{41}{841} = ( 29 ) 2 ( 21 + 20 ) ( 21 − 20 ) = 841 41
प्रश्न 25. Δ O P Q \Delta OPQ Δ O P Q में जिसका कोण P P P समकोण है, O P = 7 c m OP=7cm O P = 7 c m और O Q − P Q = 1 c m OQ-PQ=1cm O Q − P Q = 1 c m हो तो s i n Q sinQ s in Q तथा c o s Q cosQ cos Q के मान ज्ञात कीजिए। (प्रश्न बैंक)
हल— दिया है O P = 7 c m OP=7cm O P = 7 c m और O Q − P Q = 1 c m OQ-PQ=1cm O Q − P Q = 1 c m
⇒ O Q = P Q + 1 ⇒OQ=PQ+1 ⇒ O Q = P Q + 1
माना P Q = x c m PQ=x cm P Q = x c m O Q = ( x + 1 ) c m OQ=(x+1) cm O Q = ( x + 1 ) c m
पाइथागोरस प्रमेय से O P 2 + P Q 2 = O Q 2 OP^{2}+PQ^{2}=OQ^{2} O P 2 + P Q 2 = O Q 2
⇒ ( 7 ) 2 + x 2 = ( x + 1 ) 2 ⇒(7)^{2}+x^{2}=(x+1)^{2} ⇒ ( 7 ) 2 + x 2 = ( x + 1 ) 2
⇒ 49 + x 2 = x 2 + 2 x + 1 ⇒49+x^{2}=x^{2}+2x+1 ⇒ 49 + x 2 = x 2 + 2 x + 1
⇒ 2 x = 48 ⇒ x = 24 c m ⇒2x=48⇒x=24 cm ⇒ 2 x = 48 ⇒ x = 24 c m
अतः P Q = 24 , O Q = 25 PQ=24,OQ=25 P Q = 24 , O Q = 25
[Right angled triangle OPQ with P at right angle, OP=7, PQ=24, OQ=25]
∴ s i n Q = 7 25 ∴sinQ=\frac{7}{25} ∴ s in Q = 25 7 c o s Q = 24 25 cosQ=\frac{24}{25} cos Q = 25 24
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