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Class 10 Maths MCQs Chapter 8 Introduction to Trigonometry
1. The value of cos 0°. cos 1°. cos 2°. cos 3°… cos 89° cos 90° is
(a) 1
(b) -1
(c) 0
(d)
Answer
Answer: c
2. If x tan 45° sin 30° = cos 30° tan 30°, then x is equal to
(a) √3
(b)
(c)
(d) 1
Answer
Answer: d
3. If x and y are complementary angles, then
(a) sin x = sin y
(b) tan x = tan y
(c) cos x = cos y
(d) sec x = cosec y
Answer
Answer: d
4. sin 2B = 2 sin B is true when B is equal to
(a) 90°
(b) 60°
(c) 30°
(d) 0°
Answer
Answer: d
5. If A, B and C are interior angles of a ΔABC then 
Answer
Answer: a
6. If A and (2A – 45°) are acute angles such that sin A = cos (2A – 45°), then tan A is equal to
(a) 0
(b)
(c) 1
(d) √3
Answer
Answer: c
7. If y sin 45° cos 45° = tan2 45° – cos2 30°, then y = …
(a) –
(b)
(c) -2
(d) 2
Answer
Answer: b
8. If sin θ + sin² θ = 1, then cos² θ + cos4 θ = ..
(a) -1
(b) 0
(c) 1
(d) 2
Answer
Answer: c
9. 5 tan² A – 5 sec² A + 1 is equal to
(a) 6
(6) -5
(c) 1
(d) -4
Answer
Answer: d
10. If sec A + tan A = x, then sec A =
Answer
Answer: d
11. If sec A + tan A = x, then tan A =
Answer
Answer: b
Answer
Answer: b
13. If x = a cos 0 and y = b sin 0, then b2x2 + a2y2 =
(a) ab
(b) b² + a²
(c) a²b²
(d) a4b4
Answer
Answer: c
14. What is the maximum value of 
(a) 0
(b) 1
(c)
(d) 2
Answer
Answer: b
15. What is the minimum value of sin A, 0 ≤ A ≤ 90°
(a) -1
(b) 0
(c) 1
(d)
Answer
Answer: b
16. What is the minimum value of cos θ, 0 ≤ θ ≤ 90°
(a) -1
(b) 0
(c) 1
(d)
Answer
Answer: b
17. Given that sin θ = 
Answer
Answer: c
18. If cos 9A = sin A and 9A < 90°, then the value of tan 5A is
(a) 0
(b) 1
(c)
(d) √3
Answer
Answer: b
19. If in ΔABC, ∠C = 90°, then sin (A + B) =
(a) 0
(b) 1/2
(c)
(d) 1
Answer
Answer: d
20. If sin A – cos A = 0, then the value of sin4 A + cos4 A is
(a) 2
(b) 1
(c)
(d)
Answer
Answer: d
21. Ratios of sides of a right triangle with respect to its acute angles are known as
(a) trigonometric identities
(b) trigonometry
(c) trigonometric ratios of the angles
(d) none of these
Answer/Explanation
Answer: c
Explaination: (c) trigonometric ratios of the angles
22. If tan θ =
Answer/Explanation
Answer: b
Explaination:
23. Match the Columns:
(a) 1 – A, 2 – C, 3 – B
(b) 1 – B, 2 – C, 3 – A
(c) 1 – B, 2 – C, 3 – D
(d) 1 – D, 2 – B, 3 – A
Answer/Explanation
Answer: b
Explaination:
24. In the given figure, if AB = 14 cm, then the value of tan B is:
Answer/Explanation
Answer: a
Explaination:
25. Match the Columns:
(a) 1 – A, 2 – C, 3 – B
(b) 1 – C, 2 – A, 3 – D
(c) 1 – B, 2 – A, 3 – E
(d) 1 – B, 2 – D, 3 – A
Answer/Explanation
Answer: d
Explaination: (d) definition of trigonometric ratios.
26. The value of sin² 30° – cos² 30° is
Answer/Explanation
Answer:
Explaination:
27. If 3 cot θ = 2, then the value of tan θ
Answer/Explanation
Answer: b
Explaination:
3 cot θ = 2
⇒ cot θ
tan θ =
28. If ∆ABC is right angled at C, then the value of cos (A + B) is [NCERT Exemplar Problems]
Answer/Explanation
Answer: a
Explaination:
(a) ∆ ABC is right angled at C,
∴ A + B + C = 180°
A + B = 180° – 90° = 90° (∵ ∠C = 90°)
cos (A + B) = cos 90° = 0
29. If 0° < θ < 90°, then sec 0 is (a) >1
(b) < 1
(c) =1
(d) 0
Answer/Explanation
Answer: a
Explaination:
(a) ∵ sec θ =
∵ sce θ > 1.
30. If sin 0 = √3 cos θ, 0° < θ < 90°, then θ is equal to
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Answer/Explanation
Answer: c
Explaination:
31. If cos (α + β) = 0, then sin (α – β) can be reduced to [NCERT Exemplar Problems]
(a) cos β
(b) cos 2β
(c) sin α
(d) sin 2α
Answer/Explanation
Answer: b
Explaination:
(b) cos (α + β) = 0 = cos 90°
⇒ α + β = 90°
sin (α – β) = sin (α – β + β – β)
= sin (α + β – 2β)
= sin (90° – 2β) = cos 2β
32. If cos 9α = sin a and 9α < 90°, then the value of tan 5α is [NCERT Exemplar Problems]
(a)
(b) √3
(c) 1
(d) 0
Answer/Explanation
Answer: c
Explaination:
cos 9α = sin α
⇒ cos 9α = cos (90° – α)
⇒ 9α = 90° – α
⇒ 10α = 90°
⇒ α = 9°
∴ tan 5α = tan 5 × 9°
= tan 45° = 1
33. sin (45° + θ) – cos (45° – θ) is equal to
(a) 2 cos θ
(b) 0
(c) 2 sin θ
(d) 1
Answer/Explanation
Answer: b
Explaination:
sin (45° + θ) – cos (45° – θ)
= sin {90° – (45° – θ)} – cos (45° – θ)
= cos (45° – θ) – cos (45° – θ) = 0
34. The value of sin² 5° + sin² 10° + sin² 15° + … + sin² 90° is equal to
(a) 8
(b) 8.5
(c) 9
(d) 9.5
Answer/Explanation
Answer: d
Explaination:
sin² 5° + sin² 10° + sin² 15° + ….. + sin² 90°
= (sin² 5° + sin² 85°) + (sin² 10° + sin² 80°) + … + (sin² 40° + sin² 50°) + sin² 45° + sin² 90°
= (sin² 5° + cos² 5°) + (sin 10° + cos² 10°) + … + (sin² 40° + cos² 40°) + 
= 1 + 1 + 1 + … 8 times + 
= 9
35. The value of the expression [cosec (75° + θ) – sec (15° – θ) – tan (55° + θ) + cot (35° – θ)] is [NCERT Exemplar Problems]
(a) -1
(b) 0
(c) 1
(d)
Answer/Explanation
Answer: b
Explaination:
cosec (75° + θ) – sec (15° – θ) – tan (55° + θ) + cot (35° – θ)
= cosec {90° – (15° – θ)} – sec (15° – θ) – tan {90° – (35° – θ)} + cot (35° – θ)
= sec (15° – θ) – sec (15° – θ) – cot (35° – θ) + cot (35° – θ) = 0
36. sin (90° – A) =
(a) sin A
(b) tan A
(c) cos A
(d) cosec A
Answer
Answer: c
37. tan A =
Answer/Explanation
Answer: c
Explaination:
38. If cosec A – cot A = —, then cosec A =
Answer/Explanation
Answer: d
Explaination:
cosec A – cot A = 
Also cosec² A – cot² A = 1
⇒ (cosec A – cot A) (cosec A + cot A) = 1
⇒
⇒ cosec A + cot A =
From (i) and (ii), cosec A =
39. If sin x + cosec x = 2, then sin19x + cosec20x =
(a) 219
(b) 220
(c) 2
(d) 239
Answer/Explanation
Answer: c
Explaination:
(c) sin x + cosec x = 2
⇒ sin x + 
⇒ sin² x + 1 = 2 sin x
⇒ (sin x – 1)² = 0 => sin x = 1 => cosec x = 1
∴ sin19 x + cosec20 x = 1 + 1 = 2
40. If sin θ – cos θ = 0, then the value of (sin4 θ + cos4 θ) is
Answer/Explanation
Answer: c
Explaination:
sin θ – cos θ = 0
⇒ (sin θ – cos θ)² = 0
⇒ sin2²θ + cos²θ – 2 sin θ cos θ = 0
⇒ – 2 sin θ cos θ = – 1
⇒ 2 sin θ cos θ=1
⇒ sin θ cos θ =
⇒ sin²θ cos²θ =
sin4θ + cos4θ = sin4θ + cos4θ + 2 sin²θ cos²θ – 2 sin²θ cos²θ
= (sin²θ + cos² θ)² – 2 sin²θ cos²θ
= (1)² – 2 × 
41. sec A =
Answer/Explanation
Answer: d
Explaination:
42. 
(a) tan² A
(b) sec² A
(c) cosec² A – 1
(d) 1 – sin² A
Answer/Explanation
Answer: c
Explaination:
43. If sec A + tan A = x, then tan A =
Answer/Explanation
Answer: c
Explaination:
secA + tanA = x … (i)
Also sec² A – tan² A = 1
⇒ (sec A – tan A) (sec A + tan A) = 1
⇒ x (sec A – tan A)
∴ sec A – tan A = 
Now, subtracting (ii) from (i), we have
tan A =
44. Reciprocal of cot A is _____ .
Answer/Explanation
Answer:
Explaination: tan A
45. Reciprocal of cosec A is _____ .
Answer/Explanation
Answer:
Explaination: sin A
46. In ∆ABC, right angled at B, AB = 5 cm and sin C =
Answer/Explanation
Answer:
Explaination:
47. If sec θ = 
Answer/Explanation
Answer:
Explaination:
48. In ∆ABC, right angled at B, if AB = 12 cm and BC = 5 cm, find
(i) sin A and tan A, (ii) sin C and cot C.
Answer/Explanation
Answer:
Explaination:
49. If sin A = cos A, 0° < A < 90°, then A is equal to _____ .
Answer/Explanation
Answer:c
Explaination:
sin A = cos A sin A
⇒ 
⇒ tan A = 1
⇒ A = 45°
50. If sin θ1 + sin θ2 + sin θ3 = 3, 0° < θ1 θ2, θ3 ≤ 90°, then cos θ1, + cos θ2, + cos θ3 = _____ .
Answer/Explanation
Answer:
Explaination:
Hint: Maximum value of sin 0 = 1
sin θ1 + sin θ2 + sin θ3 = 3
⇒ sin θ1 = 1, sin θ2 = 1 and sin θ3 = 1
⇒ θ1 = 90°, θ2 = 90° and θ3 = 90°
∴ cos θ1 + cos θ2 + cos θ3 = 0
51. Evaluate:
sin² 60° + 2 tan 45° – cos² 30° [Allahabad 2019]
Answer/Explanation
Answer:
Explaination:
52. Given A = 30°, verify sin 2A = 2 sin A cos A.
Answer/Explanation
Answer:
Explaination:
53. If tan θ = 
Answer/Explanation
Answer:
Explaination:
54. If sin (A – B) =
Answer/Explanation
Answer:
Explaination:
sin (A – B) =
⇒ A – B = 30° ……(i)
and cos (A + B) =
⇒ A + B = 60 °…..(ii)
Solving equation (i) and (ii),
we get A = 45° and B = 15°
55. Value of 
Answer/Explanation
Answer:
Explaination:
56. If cos (40° + A) = sin 30°, the value of A is _____ .
Answer/Explanation
Answer:
Explaination:
Hint: cos (40° + A) = sin 30°
⇒ cos (40° + A) = cos (90° – 30°)
⇒ 40° + A = 60° => A = 20°
57. If tan θ = cot (30° + θ), find the value of θ.
Answer/Explanation
Answer:
Explaination:
tan θ = cot (30° + θ)
⇒ cot (90° – θ)
= cot (30° + θ)
⇒ 90° – θ = 30° + θ
⇒ 2θ = 60°
⇒ θ = 30°
58. Find the value of (sin²33° + sin²57°) [Delhi 2019]
Answer/Explanation
Answer:
Explaination:
sin²33° + sin²57°
⇒ sin²33° + sin²(90° – 33°)
⇒ sin²33° + cos²33° [Using sin(90° – θ) = cos θ]
⇒ 1 [Using sin²θ + cos² θ=1]
59. Express cot 85° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Answer/Explanation
Answer:
Explaination:
cot 85° + cos 75° = tan (90° – 85°) + sin (90° – 75°)
= tan 5° + sin 15°
60. If sec A = 
Answer/Explanation
Answer:
Explaination:
sec A =
⇒ sec(90° – B) =
⇒ cosec B =
61. If tan A + cot A = 4, then tan4 A + cot4 A =
Answer/Explanation
Answer:
Explaination:
Hint: (tan A + cot A)2 = 42
⇒ tan² A + cot² A + 2 = 16
⇒ tan² A + cot² A = 14
⇒ (tan² A + cot² A)² = (14)²
⇒ tan4 A + cot4 A + 2 = 196
tan4 A + cot4 A = 194
62. If sin x + sin²x = 1, then value of cos² x + cos4 x
Answer/Explanation
Answer:
Explaination:
Hint: sin x + sin² x = 1
⇒ sin x = 1 – sin² x
⇒ sin x = cos² x
cos² x + cos4 x = cos² x (1 + cos² x)
= sin x (1 + sin x)
= sin x + sin² x = 1
63. If tan A = 
Answer/Explanation
Answer:c
Explaination:
64. If cot θ = 
Answer/Explanation
Answer:
Explaination:
65. If sin θ = 
Answer/Explanation
Answer:
Explaination:
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