--- 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 ∘
(A) 2 \sqrt{2} 2 1 1 1 0 0 0 2 2 2\sqrt{2} 2 2 (A) 13 12 \frac{13}{12} 12 13 5 12 \frac{5}{12} 12 5 5 13 \frac{5}{13} 13 5 13 5 \frac{13}{5} 5 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
(A) 1 4 \frac{1}{4} 4 1 3 2 \frac{\sqrt{3}}{2} 2 3 1 1 1 1 2 \frac{1}{\sqrt{2}} 2 1 (A) 1 3 \frac{1}{\sqrt{3}} 3 1 1 1 1 0 0 0 3 \sqrt{3} 3
5. t a n 45 ∘ + c o t 45 ∘ tan45^{∘}+cot45^{∘} t an 4 5 ∘ + co t 4 5 ∘
(A) 1 1 1 2 2 2 3 3 3 0 0 0 (A) s e c 48 ∘ sec48^{∘} sec 4 8 ∘ t a n 48 ∘ tan48^{∘} t an 4 8 ∘ s i n 48 ∘ sin48^{∘} s in 4 8 ∘ 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
(A) c o s θ cos \theta cos θ s i n θ sin \theta s in θ s e c θ sec \theta sec θ c o t θ cot \theta co tθ (A) c o s θ cos \theta cos θ s e c θ sec \theta sec θ s i n θ sin \theta s in θ 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
(A) 2 2 2 1 1 1 1 2 \frac{1}{2} 2 1 − 1 -1 − 1 (A) 2 3 \frac{2}{\sqrt{3}} 3 2 3 2 \frac{\sqrt{3}}{2} 2 3 8 8 8 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
(A) 30 ∘ 30^{∘} 3 0 ∘ 45 ∘ 45^{∘} 4 5 ∘ 60 ∘ 60^{∘} 6 0 ∘ 90 ∘ 90^{∘} 9 0 ∘ (A) 1 2 \frac{1}{\sqrt{2}} 2 1 2 2 2 1 1 1 0 0 0
13. यदि c o s ( 90 ∘ − θ ) = 1 2 cos(90^{∘}- \theta )=\frac{1}{2} cos ( 9 0 ∘ − θ ) = 2 1 θ \theta θ
(A) 90 ∘ 90^{∘} 9 0 ∘ 60 ∘ 60^{∘} 6 0 ∘ 45 ∘ 45^{∘} 4 5 ∘ 30 ∘ 30^{∘} 3 0 ∘ (A) 2 2 2 1 1 1 1 2 \frac{1}{2} 2 1 0 0 0
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 ∘ 1 1 1 s i n 45 ∘ sin45^{∘} s in 4 5 ∘ s i n 0 ∘ sin0^{∘} s in 0 ∘ (A) 4 3 \frac{4}{3} 3 4 5 2 \frac{5}{2} 2 5 3 4 \frac{3}{4} 4 3 1 3 \frac{1}{3} 3 1
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
(A) 8 17 \frac{8}{17} 17 8 17 7 \frac{17}{7} 7 17 1 17 \frac{1}{17} 17 1 7 17 \frac{7}{17} 17 7 (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 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 a + b a − b \frac{a+b}{a-b} a − b a + b 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
(A) π 4 \frac{ \pi }{4} 4 π π 3 \frac{ \pi }{3} 3 π π 2 \frac{ \pi }{2} 2 π π 6 \frac{ \pi }{6} 6 π
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 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 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 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 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
(A) 1 (B) 9 (C) 8 (D) 0
(A) c o t 2 θ cot^{2} \theta co t 2 θ − 1 -1 − 1 (C) s e c 2 θ sec^{2} \theta se c 2 θ 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 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 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 ∘
(A) 3 (B) 1 3 \frac{1}{3} 3 1 (A) 1 2 \frac{1}{\sqrt{2}} 2 1 3 2 \frac{\sqrt{3}}{2} 2 3 1 2 \frac{1}{2} 2 1 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 )
(A) 0 (B) 1 (C) 2 (D) 3
(A) 3 4 \frac{3}{4} 4 3 4 5 \frac{4}{5} 5 4 5 4 \frac{5}{4} 4 5 5 3 \frac{5}{3} 3 5
30. 2 s i n 45 ∘ c o s 45 ∘ 2sin45^{∘}cos45^{∘} 2 s in 4 5 ∘ cos 4 5 ∘
( माध्य. शिक्षा बोर्ड, 2024 )
(A) 2 (B) 0 (C) 1 2 \frac{1}{2} 2 1 (A) 0 (B) 1 (C) 2 (D) 3
32. s i n 45 ∘ c o s 45 ∘ \frac{sin45^{∘}}{cos45^{∘}} cos 4 5 ∘ s in 4 5 ∘
(A) 0 (B) 1 (C) 2 (D) 3
(A) 12 13 \frac{12}{13} 13 12 12 5 \frac{12}{5} 5 12 13 5 \frac{13}{5} 5 13 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 )
(A) 9 49 \frac{9}{49} 49 9 9 41 \frac{9}{41} 41 9 49 9 \frac{49}{9} 9 49 41 9 \frac{41}{9} 9 41 (A) 5 (B) 3 (C) 4 3 \frac{4}{3} 3 4
36. यदि t a n 3 x = 1 tan3x=1 t an 3 x = 1 x x x
( प्रश्न बैंक )
(A) 15 ∘ 15^{∘} 1 5 ∘ 30 ∘ 30^{∘} 3 0 ∘ 45 ∘ 45^{∘} 4 5 ∘ 90 ∘ 90^{∘} 9 0 ∘ (A) 3 4 \frac{3}{4} 4 3 4 5 \frac{4}{5} 5 4 5 4 \frac{5}{4} 4 5 5 3 \frac{5}{3} 3 5
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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