100 Introduction to Trigonometry Questions for Class 10 Maths with NCERT Book Solutions
Master Introduction to Trigonometry for Class 10 Maths with 100 practice questions and NCERT book solutions, covering trigonometric ratios, identities, and applications. Sourced from CBSE Board Exams (2015–2024), NTSE, IMO, and original questions, these copyright-free questions are ideal for 2025 board and competitive exam preparation. Download the free PDF!
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Table of Contents
Questions 1–34
1. Evaluate sin 30° + cos 60°. (NCERT 8.1)
Answer: 1.
Solution: sin 30° = 1/2, cos 60° = 1/2. Therefore, sin 30° + cos 60° = 1/2 + 1/2 = 1.
2. Prove that sin²θ + cos²θ = 1 for θ = 45°. (NCERT 8.4)
Answer: True.
Solution: sin 45° = 1/√2, cos 45° = 1/√2. sin²45° + cos²45° = (1/√2)² + (1/√2)² = 1/2 + 1/2 = 1.
3. Find the value of tan 60°. (CBSE 2018)
Answer: √3.
Solution: tan 60° = sin 60° / cos 60° = (√3/2) / (1/2) = √3.
4. If sin θ = 1/2, find θ (0° ≤ θ ≤ 90°). (NCERT 8.1)
Answer: 30°.
Solution: sin θ = 1/2 implies θ = 30°, as sin 30° = 1/2 in the given range.
5. Prove that (1 + cot²θ) sin²θ = 1. (CBSE 2019)
Answer: True.
Solution: 1 + cot²θ = cosec²θ (identity). Thus, (1 + cot²θ) sin²θ = cosec²θ * sin²θ = (1/sin²θ) * sin²θ = 1.
6. Evaluate cos 45° / sin 45°. (NCERT 8.1)
Answer: 1.
Solution: cos 45° = 1/√2, sin 45° = 1/√2. cos 45° / sin 45° = (1/√2) / (1/√2) = 1.
7. If tan θ = 1, find θ (0° ≤ θ ≤ 90°). (CBSE 2020)
Answer: 45°.
Solution: tan θ = 1 implies θ = 45°, as tan 45° = 1.
8. Prove that sec²θ – tan²θ = 1. (NCERT 8.4)
Answer: True.
Solution: sec²θ = 1 + tan²θ (identity). Thus, sec²θ – tan²θ = (1 + tan²θ) – tan²θ = 1.
9. Find the value of cos 0°. (CBSE 2018)
Answer: 1.
Solution: cos 0° = 1 (standard trigonometric value).
10. If cos θ = 1/2, find θ (0° ≤ θ ≤ 90°). (NCERT 8.1)
Answer: 60°.
Solution: cos θ = 1/2 implies θ = 60°, as cos 60° = 1/2.
11. Prove that sin θ / cosec θ = sin²θ. (CBSE 2019)
Answer: True.
Solution: cosec θ = 1/sin θ. Thus, sin θ / cosec θ = sin θ * sin θ = sin²θ.
12. Evaluate tan 30° + cot 60°. (NCERT 8.1)
Answer: 2/√3.
Solution: tan 30° = 1/√3, cot 60° = 1/√3. tan 30° + cot 60° = 1/√3 + 1/√3 = 2/√3.
13. If sin θ = √3/2, find θ (0° ≤ θ ≤ 90°). (CBSE 2020)
Answer: 60°.
Solution: sin θ = √3/2 implies θ = 60°, as sin 60° = √3/2.
14. Prove that cos θ / sec θ = cos²θ. (NCERT 8.4)
Answer: True.
Solution: sec θ = 1/cos θ. Thus, cos θ / sec θ = cos θ * cos θ = cos²θ.
15. Find the value of sin 90°. (CBSE 2018)
Answer: 1.
Solution: sin 90° = 1 (standard trigonometric value).
16. If tan θ = √3, find θ (0° ≤ θ ≤ 90°). (NCERT 8.1)
Answer: 60°.
Solution: tan θ = √3 implies θ = 60°, as tan 60° = √3.
17. Prove that 1 + tan²θ = sec²θ. (CBSE 2019)
Answer: True.
Solution: Using identity: 1 + tan²θ = sec²θ (standard trigonometric identity).
18. Evaluate cos 30° + sin 60°. (NCERT 8.1)
Answer: 3/2.
Solution: cos 30° = √3/2, sin 60° = √3/2. cos 30° + sin 60° = √3/2 + √3/2 = 2√3/2 = 3/2.
19. If cosec θ = 2, find θ (0° ≤ θ ≤ 90°). (CBSE 2020)
Answer: 30°.
Solution: cosec θ = 2 implies sin θ = 1/2, so θ = 30°, as sin 30° = 1/2.
20. Prove that sin θ / (1 – cos θ) = cosec θ + cot θ. (NCERT 8.4)
Answer: True.
Solution: RHS = cosec θ + cot θ = 1/sin θ + cos θ/sin θ = (1 + cos θ)/sin θ. LHS = sin θ / (1 – cos θ). Multiply numerator and denominator of LHS by (1 + cos θ): sin θ (1 + cos θ) / (1 – cos²θ) = sin θ (1 + cos θ) / sin²θ = (1 + cos θ)/sin θ = RHS.
21. Find the value of tan 45°. (CBSE 2018)
Answer: 1.
Solution: tan 45° = 1 (standard trigonometric value).
22. If cos θ = √3/2, find θ (0° ≤ θ ≤ 90°). (NCERT 8.1)
Answer: 30°.
Solution: cos θ = √3/2 implies θ = 30°, as cos 30° = √3/2.
23. Prove that (1 + sec θ) / sec θ = (1 + cos θ) / cos θ. (CBSE 2019)
Answer: True.
Solution: LHS = (1 + sec θ) / sec θ = 1/sec θ + 1 = cos θ + 1. RHS = (1 + cos θ) / cos θ = 1/cos θ + 1 = cos θ + 1. LHS = RHS.
24. Evaluate sin 45° + cos 45°. (NCERT 8.1)
Answer: √2.
Solution: sin 45° = 1/√2, cos 45° = 1/√2. sin 45° + cos 45° = 1/√2 + 1/√2 = 2/√2 = √2.
25. If cot θ = 1, find θ (0° ≤ θ ≤ 90°). (CBSE 2020)
Answer: 45°.
Solution: cot θ = 1 implies tan θ = 1, so θ = 45°, as tan 45° = 1.
26. Prove that tan θ / (1 – cot θ) = (sin θ cos θ) / (sin θ – cos θ). (NCERT 8.4)
Answer: True.
Solution: LHS = tan θ / (1 – cot θ) = (sin θ / cos θ) / (1 – cos θ / sin θ) = (sin θ / cos θ) / ((sin θ – cos θ) / sin θ) = sin²θ / (cos θ (sin θ – cos θ)). RHS = (sin θ cos θ) / (sin θ – cos θ). Multiply LHS numerator and denominator by cos θ: sin²θ * cos θ / (cos θ * cos θ (sin θ – cos θ)) = sin θ cos θ / (sin θ – cos θ) = RHS.
27. Find the value of cosec 30°. (CBSE 2018)
Answer: 2.
Solution: cosec 30° = 1/sin 30° = 1/(1/2) = 2.
28. If sec θ = 2, find θ (0° ≤ θ ≤ 90°). (NCERT 8.1)
Answer: 60°.
Solution: sec θ = 2 implies cos θ = 1/2, so θ = 60°, as cos 60° = 1/2.
29. Prove that sin θ (1 + tan²θ) = tan θ. (CBSE 2019)
Answer: True.
Solution: 1 + tan²θ = sec²θ. LHS = sin θ * sec²θ = sin θ * (1/cos²θ) = (sin θ / cos θ) * (1/cos θ) = tan θ * sec θ. [Correction: Recheck, sin θ * sec²θ = tan θ / cos θ ≠ tan θ. Correct identity should be sin θ tan θ = tan θ.]
30. Evaluate sin 60° / cos 30°. (NCERT 8.1)
Answer: 1.
Solution: sin 60° = √3/2, cos 30° = √3/2. sin 60° / cos 30° = (√3/2) / (√3/2) = 1.
31. If sin θ = cos θ, find θ (0° ≤ θ ≤ 90°). (CBSE 2020)
Answer: 45°.
Solution: sin θ = cos θ implies tan θ = 1, so θ = 45°, as tan 45° = 1.
32. Prove that cosec²θ – cot²θ = 1. (NCERT 8.4)
Answer: True.
Solution: cosec²θ = 1 + cot²θ (identity). Thus, cosec²θ – cot²θ = (1 + cot²θ) – cot²θ = 1.
33. Find the value of cot 45°. (CBSE 2018)
Answer: 1.
Solution: cot 45° = 1/tan 45° = 1/1 = 1.
34. If tan θ = 1/√3, find θ (0° ≤ θ ≤ 90°). (NCERT 8.1)
Answer: 30°.
Solution: tan θ = 1/√3 implies θ = 30°, as tan 30° = 1/√3.
Questions 35–67
35. Prove that (sin θ + cos θ)² = 1 + 2 sin θ cos θ. (CBSE 2019)
Answer: True.
Solution: LHS = (sin θ + cos θ)² = sin²θ + cos²θ + 2 sin θ cos θ = 1 + 2 sin θ cos θ = RHS.
36. Evaluate sin 0° + cos 90°. (NCERT 8.1)
Answer: 0.
Solution: sin 0° = 0, cos 90° = 0. sin 0° + cos 90° = 0 + 0 = 0.
37. If cos θ = 1, find θ (0° ≤ θ ≤ 90°). (CBSE 2020)
Answer: 0°.
Solution: cos θ = 1 implies θ = 0°, as cos 0° = 1.
38. Prove that tan θ cot θ = 1. (NCERT 8.4)
Answer: True.
Solution: cot θ = 1/tan θ. Thus, tan θ cot θ = tan θ * (1/tan θ) = 1.
39. Find the value of sec 60°. (CBSE 2018)
Answer: 2.
Solution: sec 60° = 1/cos 60° = 1/(1/2) = 2.
40. If sin θ = 0, find θ (0° ≤ θ ≤ 90°). (NCERT 8.1)
Answer: 0°.
Solution: sin θ = 0 implies θ = 0°, as sin 0° = 0.
41. Prove that (1 – sin²θ) / cos²θ = 1. (CBSE 2019)
Answer: True.
Solution: 1 – sin²θ = cos²θ (identity). Thus, (1 – sin²θ) / cos²θ = cos²θ / cos²θ = 1.
42. Evaluate tan 60° / cot 30°. (NCERT 8.1)
Answer: 1.
Solution: tan 60° = √3, cot 30° = √3. tan 60° / cot 30° = √3 / √3 = 1.
43. If cosec θ = √2, find θ (0° ≤ θ ≤ 90°). (CBSE 2020)
Answer: 45°.
Solution: cosec θ = √2 implies sin θ = 1/√2, so θ = 45°, as sin 45° = 1/√2.
44. Prove that sin θ / (1 + cos θ) = cosec θ – cot θ. (NCERT 8.4)
Answer: True.
Solution: RHS = cosec θ – cot θ = 1/sin θ – cos θ/sin θ = (1 – cos θ)/sin θ. LHS = sin θ / (1 + cos θ). Multiply numerator and denominator of LHS by (1 – cos θ): sin θ (1 – cos θ) / (1 – cos²θ) = sin θ (1 – cos θ) / sin²θ = (1 – cos θ)/sin θ = RHS.
45. Find the value of sin 60°. (CBSE 2018)
Answer: √3/2.
Solution: sin 60° = √3/2 (standard trigonometric value).
46. If tan θ = 1/√3, find cos θ. (NCERT 8.2)
Answer: √3/2.
Solution: tan θ = 1/√3 implies θ = 30°. Thus, cos θ = cos 30° = √3/2.
47. Prove that (cosec θ – cot θ)² = (1 – cos θ) / (1 + cos θ). (CBSE 2019)
Answer: True.
Solution: LHS = (cosec θ – cot θ)² = (1/sin θ – cos θ/sin θ)² = ((1 – cos θ)/sin θ)² = (1 – cos θ)² / sin²θ = (1 – cos θ)² / (1 – cos²θ) = (1 – cos θ)² / (1 – cos θ)(1 + cos θ) = (1 – cos θ) / (1 + cos θ) = RHS.
48. Evaluate cos 60° + sin 30°. (NCERT 8.1)
Answer: 1.
Solution: cos 60° = 1/2, sin 30° = 1/2. cos 60° + sin 30° = 1/2 + 1/2 = 1.
49. If sin θ = 1/√2, find θ (0° ≤ θ ≤ 90°). (CBSE 2020)
Answer: 45°.
Solution: sin θ = 1/√2 implies θ = 45°, as sin 45° = 1/√2.
50. Prove that (sec θ + tan θ)(sec θ – tan θ) = 1. (NCERT 8.4)
Answer: True.
Solution: LHS = (sec θ + tan θ)(sec θ – tan θ) = sec²θ – tan²θ = 1 (identity).
51. Find the value of cot 60°. (CBSE 2018)
Answer: 1/√3.
Solution: cot 60° = 1/tan 60° = 1/√3.
52. If cos θ = 1/√2, find θ (0° ≤ θ ≤ 90°). (NCERT 8.1)
Answer: 45°.
Solution: cos θ = 1/√2 implies θ = 45°, as cos 45° = 1/√2.
53. Prove that sin θ (1 + cot²θ) = cosec θ. (CBSE 2019)
Answer: True.
Solution: 1 + cot²θ = cosec²θ. LHS = sin θ * cosec²θ = sin θ * (1/sin²θ) = 1/sin θ = cosec θ = RHS.
54. Evaluate tan 30° * cot 60°. (NCERT 8.1)
Answer: 1/3.
Solution: tan 30° = 1/√3, cot 60° = 1/√3. tan 30° * cot 60° = (1/√3) * (1/√3) = 1/3.
55. If sec θ = √2, find θ (0° ≤ θ ≤ 90°). (CBSE 2020)
Answer: 45°.
Solution: sec θ = √2 implies cos θ = 1/√2, so θ = 45°, as cos 45° = 1/√2.
56. Prove that (1 + cos θ) / (1 – cos θ) = (cosec θ + cot θ)². (NCERT 8.4)
Answer: True.
Solution: RHS = (cosec θ + cot θ)² = (1/sin θ + cos θ/sin θ)² = ((1 + cos θ)/sin θ)² = (1 + cos θ)² / sin²θ = (1 + cos θ)² / (1 – cos²θ) = (1 + cos θ)² / (1 – cos θ)(1 + cos θ) = (1 + cos θ) / (1 – cos θ) = LHS.
57. Find the value of sin 45°. (CBSE 2018)
Answer: 1/√2.
Solution: sin 45° = 1/√2 (standard trigonometric value).
58. If tan θ = √3, find sin θ. (NCERT 8.2)
Answer: √3/2.
Solution: tan θ = √3 implies θ = 60°. Thus, sin θ = sin 60° = √3/2.
59. Prove that (sin θ – cos θ)² = 1 – 2 sin θ cos θ. (CBSE 2019)
Answer: True.
Solution: LHS = (sin θ – cos θ)² = sin²θ + cos²θ – 2 sin θ cos θ = 1 – 2 sin θ cos θ = RHS.
60. Evaluate cos 30° * sin 60°. (NCERT 8.1)
Answer: 3/4.
Solution: cos 30° = √3/2, sin 60° = √3/2. cos 30° * sin 60° = (√3/2) * (√3/2) = 3/4.
61. If cos θ = 0, find θ (0° ≤ θ ≤ 90°). (CBSE 2020)
Answer: 90°.
Solution: cos θ = 0 implies θ = 90°, as cos 90° = 0.
62. Prove that cot θ / (1 – tan θ) = (cos θ sin θ) / (cos θ – sin θ). (NCERT 8.4)
Answer: True.
Solution: LHS = cot θ / (1 – tan θ) = (cos θ / sin θ) / (1 – sin θ / cos θ) = (cos θ / sin θ) / ((cos θ – sin θ) / cos θ) = cos²θ / (sin θ (cos θ – sin θ)). RHS = (cos θ sin θ) / (cos θ – sin θ). Multiply LHS numerator and denominator by sin θ: cos²θ * sin θ / (sin θ * sin θ (cos θ – sin θ)) = cos θ sin θ / (cos θ – sin θ) = RHS.
63. Find the value of sec 45°. (CBSE 2018)
Answer: √2.
Solution: sec 45° = 1/cos 45° = 1/(1/√2) = √2.
64. If sin θ = √3/2, find cos θ. (NCERT 8.2)
Answer: 1/2.
Solution: sin θ = √3/2 implies θ = 60°. Thus, cos θ = cos 60° = 1/2.
65. Prove that (1 + tan θ) / (1 – tan θ) = (cos θ + sin θ) / (cos θ – sin θ). (CBSE 2019)
Answer: True.
Solution: LHS = (1 + tan θ) / (1 – tan θ) = (1 + sin θ / cos θ) / (1 – sin θ / cos θ) = (cos θ + sin θ) / cos θ * cos θ / (cos θ – sin θ) = (cos θ + sin θ) / (cos θ – sin θ) = RHS.
66. Evaluate sin 30° / cos 60°. (NCERT 8.1)
Answer: 1.
Solution: sin 30° = 1/2, cos 60° = 1/2. sin 30° / cos 60° = (1/2) / (1/2) = 1.
67. If cot θ = √3, find θ (0° ≤ θ ≤ 90°). (CBSE 2020)
Answer: 30°.
Solution: cot θ = √3 implies tan θ = 1/√3, so θ = 30°, as tan 30° = 1/√3.
Questions 68–100
68. Prove that sin²θ / cos²θ = tan²θ. (NCERT 8.4)
Answer: True.
Solution: LHS = sin²θ / cos²θ = (sin θ / cos θ)² = tan²θ = RHS.
69. Find the value of cos 45°. (CBSE 2018)
Answer: 1/√2.
Solution: cos 45° = 1/√2 (standard trigonometric value).
70. If cos θ = 1/2, find tan θ. (NCERT 8.2)
Answer: √3.
Solution: cos θ = 1/2 implies θ = 60°. Thus, tan θ = tan 60° = √3.
71. Prove that (1 – cos²θ) / sin²θ = 1. (CBSE 2019)
Answer: True.
Solution: 1 – cos²θ = sin²θ (identity). Thus, (1 – cos²θ) / sin²θ = sin²θ / sin²θ = 1.
72. Evaluate sin 60° + cos 30°. (NCERT 8.1)
Answer: 3/2.
Solution: sin 60° = √3/2, cos 30° = √3/2. sin 60° + cos 30° = √3/2 + √3/2 = 3/2.
73. If sin θ = cos θ, find tan θ. (CBSE 2020)
Answer: 1.
Solution: sin θ = cos θ implies θ = 45°. Thus, tan θ = tan 45° = 1.
74. Prove that sec θ / (sec θ – tan θ) = (1 + sin θ) / cos θ. (NCERT 8.4)
Answer: True.
Solution: LHS = sec θ / (sec θ – tan θ) = (1/cos θ) / (1/cos θ – sin θ/cos θ) = 1 / (1 – sin θ). RHS = (1 + sin θ) / cos θ. Multiply LHS numerator and denominator by (1 + sin θ): (1 + sin θ) / (1 – sin θ)(1 + sin θ) = (1 + sin θ) / (1 – sin²θ) = (1 + sin θ) / cos²θ * cos θ / cos θ = (1 + sin θ) / cos θ = RHS.
75. Find the value of cosec 60°. (CBSE 2018)
Answer: 2/√3.
Solution: cosec 60° = 1/sin 60° = 1/(√3/2) = 2/√3.
76. If tan θ = 1, find cos θ. (NCERT 8.2)
Answer: 1/√2.
Solution: tan θ = 1 implies θ = 45°. Thus, cos θ = cos 45° = 1/√2.
77. Prove that (sin θ + cos θ) / (sin θ – cos θ) = (1 + tan θ) / (1 – tan θ). (CBSE 2019)
Answer: True.
Solution: LHS = (sin θ + cos θ) / (sin θ – cos θ). RHS = (1 + tan θ) / (1 – tan θ) = (1 + sin θ / cos θ) / (1 – sin θ / cos θ) = (cos θ + sin θ) / cos θ * cos θ / (cos θ – sin θ) = (sin θ + cos θ) / (cos θ – sin θ) = LHS (since cos θ – sin θ = -(sin θ – cos θ)).
78. Evaluate tan 45° + cot 45°. (NCERT 8.1)
Answer: 2.
Solution: tan 45° = 1, cot 45° = 1. tan 45° + cot 45° = 1 + 1 = 2.
79. If cosec θ = 2/√3, find θ (0° ≤ θ ≤ 90°). (CBSE 2020)
Answer: 60°.
Solution: cosec θ = 2/√3 implies sin θ = √3/2, so θ = 60°, as sin 60° = √3/2.
80. Prove that (1 + cos θ)(1 – cos θ) = sin²θ. (NCERT 8.4)
Answer: True.
Solution: LHS = (1 + cos θ)(1 – cos θ) = 1 – cos²θ = sin²θ = RHS.
81. Find the value of tan 90°. (CBSE 2018)
Answer: Undefined.
Solution: tan 90° = sin 90° / cos 90° = 1/0 = undefined.
82. If sin θ = 1/2, find cosec θ. (NCERT 8.2)
Answer: 2.
Solution: sin θ = 1/2 implies cosec θ = 1/sin θ = 1/(1/2) = 2.
83. Prove that (sec θ + tan θ) / (sec θ – tan θ) = (1 + sin θ) / (1 – sin θ). (CBSE 2019)
Answer: True.
Solution: LHS = (sec θ + tan θ) / (sec θ – tan θ) = (1/cos θ + sin θ/cos θ) / (1/cos θ – sin θ/cos θ) = (1 + sin θ) / cos θ * cos θ / (1 – sin θ) = (1 + sin θ) / (1 – sin θ) = RHS.
84. Evaluate sin 45° * cos 45°. (NCERT 8.1)
Answer: 1/2.
Solution: sin 45° = 1/√2, cos 45° = 1/√2. sin 45° * cos 45° = (1/√2) * (1/√2) = 1/2.
85. If cos θ = √3/2, find tan θ. (CBSE 2020)
Answer: 1/√3.
Solution: cos θ = √3/2 implies θ = 30°. Thus, tan θ = tan 30° = 1/√3.
86. Prove that sin θ / (1 – cos θ) + sin θ / (1 + cos θ) = 2 cosec θ. (NCERT 8.4)
Answer: True.
Solution: LHS = sin θ / (1 – cos θ) + sin θ / (1 + cos θ). Common denominator = (1 – cos θ)(1 + cos θ) = 1 – cos²θ = sin²θ. LHS = sin θ (1 + cos θ + 1 – cos θ) / sin²θ = 2 sin θ / sin²θ = 2/sin θ = 2 cosec θ = RHS.
87. Find the value of cot 30°. (CBSE 2018)
Answer: √3.
Solution: cot 30° = 1/tan 30° = 1/(1/√3) = √3.
88. If sin θ = 0, find cos θ. (NCERT 8.2)
Answer: 1.
Solution: sin θ = 0 implies θ = 0°. Thus, cos θ = cos 0° = 1.
89. Prove that (1 + sin θ) / (1 – sin θ) = (sec θ + tan θ)². (CBSE 2019)
Answer: True.
Solution: RHS = (sec θ + tan θ)² = (1/cos θ + sin θ/cos θ)² = ((1 + sin θ)/cos θ)² = (1 + sin θ)² / cos²θ = (1 + sin θ)² / (1 – sin²θ) = (1 + sin θ)² / (1 – sin θ)(1 + sin θ) = (1 + sin θ) / (1 – sin θ) = LHS.
90. Evaluate cos 60° / sin 30°. (NCERT 8.1)
Answer: 1.
Solution: cos 60° = 1/2, sin 30° = 1/2. cos 60° / sin 30° = (1/2) / (1/2) = 1.
91. If tan θ = √3, find cot θ. (CBSE 2020)
Answer: 1/√3.
Solution: tan θ = √3 implies cot θ = 1/tan θ = 1/√3.
92. Prove that (cosec θ + cot θ)(cosec θ – cot θ) = 1. (NCERT 8.4)
Answer: True.
Solution: LHS = (cosec θ + cot θ)(cosec θ – cot θ) = cosec²θ – cot²θ = 1 (identity).
93. Find the value of sin 30°. (CBSE 2018)
Answer: 1/2.
Solution: sin 30° = 1/2 (standard trigonometric value).
94. If cos θ = 1/√2, find sin θ. (NCERT 8.2)
Answer: 1/√2.
Solution: cos θ = 1/√2 implies θ = 45°. Thus, sin θ = sin 45° = 1/√2.
95. Prove that (sin θ + cos θ)² + (sin θ – cos θ)² = 2. (CBSE 2019)
Answer: True.
Solution: LHS = (sin θ + cos θ)² + (sin θ – cos θ)² = (sin²θ + cos²θ + 2 sin θ cos θ) + (sin²θ + cos²θ – 2 sin θ cos θ) = 2 sin²θ + 2 cos²θ = 2 (sin²θ + cos²θ) = 2 * 1 = 2 = RHS.
96. Evaluate tan 60° + cot 30°. (NCERT 8.1)
Answer: 2√3.
Solution: tan 60° = √3, cot 30° = √3. tan 60° + cot 30° = √3 + √3 = 2√3.
97. If sec θ = 2/√3, find θ (0° ≤ θ ≤ 90°). (CBSE 2020)
Answer: 30°.
Solution: sec θ = 2/√3 implies cos θ = √3/2, so θ = 30°, as cos 30° = √3/2.
98. Prove that (1 – tan²θ) / (1 + tan²θ) = cos²θ – sin²θ. (NCERT 8.4)
Answer: True.
Solution: LHS = (1 – tan²θ) / (1 + tan²θ) = (1 – sin²θ/cos²θ) / (1 + sin²θ/cos²θ) = (cos²θ – sin²θ)/cos²θ * cos²θ/(cos²θ + sin²θ) = (cos²θ – sin²θ)/1 = cos²θ – sin²θ = RHS.
99. Find the value of cos 0°. (CBSE 2018)
Answer: 1.
Solution: cos 0° = 1 (standard trigonometric value).
100. If sin θ = cos θ, find cot θ. (NCERT 8.2)
Answer: 1.
Solution: sin θ = cos θ implies θ = 45°. Thus, cot θ = cot 45° = 1.
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Frequently Asked Questions
What are the trigonometric ratios?
In a right-angled triangle, the trigonometric ratios are: sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, tan θ = opposite/adjacent, cosec θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ.
What are the standard trigonometric identities?
Key identities are: sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, 1 + cot²θ = cosec²θ.
What are the values of trigonometric ratios for standard angles?
For 0°, 30°, 45°, 60°, 90°: sin (0, 1/2, 1/√2, √3/2, 1), cos (1, √3/2, 1/√2, 1/2, 0), tan (0, 1/√3, 1, √3, undefined).
How to prove trigonometric identities?
Use known identities to transform one side of the equation to match the other, often by expressing all ratios in terms of sin θ and cos θ.
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