Categories: MCQ Questions

# MCQ Questions for Class 10 Maths Introduction to Trigonometry with Answers

Free PDF Download of CBSE Class 10 Maths Chapter 8 Introduction to Trigonometry Multiple Choice Questions with Answers. MCQ Questions for Class 10 Maths with Answers was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 10 Maths Introduction to Trigonometry MCQs with Answers to know their preparation level.

## 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) $$\frac{1}{\sqrt{2}}$$

2. If x tan 45° sin 30° = cos 30° tan 30°, then x is equal to
(a) √3
(b) $$\frac{1}{2}$$
(c) $$\frac{1}{\sqrt{2}}$$
(d) 1

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

4. sin 2B = 2 sin B is true when B is equal to
(a) 90°
(b) 60°
(c) 30°
(d) 0°

5. If A, B and C are interior angles of a ΔABC then $$\cos \left(\frac{\mathrm{B}+\mathrm{C}}{2}\right)$$ is equal to

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) $$\frac{1}{\sqrt{3}}$$
(c) 1
(d) √3

7. If y sin 45° cos 45° = tan2 45° – cos2 30°, then y = …
(a) –$$\frac{1}{2}$$
(b) $$\frac{1}{2}$$
(c) -2
(d) 2

8. If sin θ + sin² θ = 1, then cos² θ + cos4 θ = ..
(a) -1
(b) 0
(c) 1
(d) 2

9. 5 tan² A – 5 sec² A + 1 is equal to
(a) 6
(6) -5
(c) 1
(d) -4

10. If sec A + tan A = x, then sec A =

11. If sec A + tan A = x, then tan A =

13. If x = a cos 0 and y = b sin 0, then b2x2 + a2y2 =
(a) ab
(b) b² + a²
(c) a²b²
(d) a4b4

14. What is the maximum value of $$\frac{1}{\csc A}$$?
(a) 0
(b) 1
(c) $$\frac{1}{2}$$
(d) 2

15. What is the minimum value of sin A, 0 ≤ A ≤ 90°
(a) -1
(b) 0
(c) 1
(d) $$\frac{1}{2}$$

16. What is the minimum value of cos θ, 0 ≤ θ ≤ 90°
(a) -1
(b) 0
(c) 1
(d) $$\frac{1}{2}$$

17. Given that sin θ = $$\frac{a}{b}$$ , then tan θ =

18. If cos 9A = sin A and 9A < 90°, then the value of tan 5A is
(a) 0
(b) 1
(c) $$\frac{1}{\sqrt{3}}$$
(d) √3

19. If in ΔABC, ∠C = 90°, then sin (A + B) =
(a) 0
(b) 1/2
(c) $$\frac{1}{\sqrt{2}}$$
(d) 1

20. If sin A – cos A = 0, then the value of sin4 A + cos4 A is
(a) 2
(b) 1
(c) $$\frac{3}{4}$$
(d) $$\frac{1}{2}$$

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

Explaination: (c) trigonometric ratios of the angles

22. If tan θ = $$\frac{a}{b}$$ then the value of

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

Explaination:

24. In the given figure, if AB = 14 cm, then the value of tan B is:

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

Explaination: (d) definition of trigonometric ratios.

26. The value of sin² 30° – cos² 30° is

Explaination:

27. If 3 cot θ = 2, then the value of tan θ

Explaination:
3 cot θ = 2
⇒ cot θ $$\frac{2}{3}$$
tan θ = $$\frac{3}{2}$$

28. If ∆ABC is right angled at C, then the value of cos (A + B) is [NCERT Exemplar Problems]

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

Explaination:
(a) ∵ sec θ = $$\frac{1}{\cos \theta}$$
∵ 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°

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

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) $$\frac{1}{\sqrt{3}}$$
(b) √3
(c) 1
(d) 0

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

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

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°) + $$\left(\frac{1}{\sqrt{2}}\right)^{2}$$ + 1
= 1 + 1 + 1 + … 8 times + $$\frac{1}{2}$$ + 1
= 9$$\frac{1}{2}$$ = 9.5

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) $$\frac{3}{2}$$

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

37. tan A =

Explaination:

38. If cosec A – cot A = —, then cosec A =

Explaination:
cosec A – cot A = $$\frac{4}{5}$$ ……(i)
Also cosec² A – cot² A = 1
⇒ (cosec A – cot A) (cosec A + cot A) = 1
⇒ $$\frac{4}{5}$$(cosec A + cot A) = 1
⇒ cosec A + cot A = $$\frac{4}{5}$$ …(ii)
From (i) and (ii), cosec A = $$\frac{41}{40}$$

39. If sin x + cosec x = 2, then sin19x + cosec20x =
(a) 219
(b) 220
(c) 2
(d) 239

Explaination:
(c) sin x + cosec x = 2
⇒ sin x + $$\frac{1}{sin x}$$ = 2
⇒ 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

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 = $$\frac{1}{2}$$
⇒ sin²θ cos²θ = $$\frac{1}{4}$$
sin4θ + cos4θ = sin4θ + cos4θ + 2 sin²θ cos²θ – 2 sin²θ cos²θ
= (sin²θ + cos² θ)² – 2 sin²θ cos²θ
= (1)² – 2 × $$\frac{1}{4}$$ = 1 – $$\frac{1}{2}$$ = $$\frac{1}{2}$$

41. sec A =

Explaination:

42.
(a) tan² A
(b) sec² A
(c) cosec² A – 1
(d) 1 – sin² A

Explaination:

43. If sec A + tan A = x, then tan A =

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 = $$\frac{1}{x}$$ ….. (ii)
Now, subtracting (ii) from (i), we have
tan A = $$\frac{x^{2}-1}{2 x}$$

44. Reciprocal of cot A is _____ .

Explaination: tan A

45. Reciprocal of cosec A is _____ .

Explaination: sin A

46. In ∆ABC, right angled at B, AB = 5 cm and sin C = $$\frac{1}{2}$$. Determine the length of side AC.

Explaination:

47. If sec θ = $$\frac{25}{7}$$, find the values of tan θ and cosec θ.

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.

Explaination:

49. If sin A = cos A, 0° < A < 90°, then A is equal to _____ .

Explaination:
sin A = cos A sin A
⇒ $$\frac{\sin A}{\cos A}$$ = 1
⇒ 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 = _____ .

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]

Explaination:

52. Given A = 30°, verify sin 2A = 2 sin A cos A.

Explaination:

53. If tan θ = $$\frac{1}{\sqrt{3}}$$ =r, then evaluate

Explaination:

54. If sin (A – B) = $$\frac{1}{2}$$, cos (A + B) = $$\frac{1}{2}$$, find A and B.

Explaination:
sin (A – B) = $$\frac{1}{2}$$
⇒ A – B = 30° ……(i)
and cos (A + B) = $$\frac{1}{2}$$
⇒ A + B = 60 °…..(ii)
Solving equation (i) and (ii),
we get A = 45° and B = 15°

55. Value of $$\frac{\tan 65^{\circ}}{\cot 25^{\circ}}$$ = _____.

Explaination:

56. If cos (40° + A) = sin 30°, the value of A is _____ .

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.

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]

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°.

Explaination:
cot 85° + cos 75° = tan (90° – 85°) + sin (90° – 75°)
= tan 5° + sin 15°

60. If sec A = $$\frac{15}{7}$$ and A + B = 90°, find the value of cosec B.

Explaination:
sec A = $$\frac{15}{7}$$
⇒ sec(90° – B) = $$\frac{15}{7}$$ [∵ A + B = 90° ⇒ A = 90°-B]
⇒ cosec B = $$\frac{15}{7}$$ [∵ sec (90° – θ) = cosec θ]

61. If tan A + cot A = 4, then tan4 A + cot4 A =

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

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 = $$\frac{5}{12}$$, find the value of 12 (sin A + cos A).sec A.

Explaination:

64. If cot θ = $$\frac{7}{8}$$, evaluate

Explaination:

65. If sin θ = $$\frac{1}{3}$$, then find the value of (2 cot² θ + 2)

Explaination:

We hope the given MCQ Questions for Class 10 Maths Introduction to Trigonometry with Answers will help you. If you have any query regarding CBSE Class 10 Maths Chapter 8 Introduction to Trigonometry Multiple Choice Questions with Answers, drop a comment below and we will get back to you at the earliest.

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