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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) \(\frac{1}{\sqrt{2}}\)

Answer

Answer: c


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

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 \(\cos \left(\frac{\mathrm{B}+\mathrm{C}}{2}\right)\) is equal to

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) \(\frac{1}{\sqrt{3}}\)
(c) 1
(d) √3

Answer

Answer: c


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

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 \(\frac{1}{\csc A}\)?
(a) 0
(b) 1
(c) \(\frac{1}{2}\)
(d) 2

Answer

Answer: b


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

Answer

Answer: b


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

Answer

Answer: b


17. Given that sin θ = \(\frac{a}{b}\) , then tan 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) \(\frac{1}{\sqrt{3}}\)
(d) √3

Answer

Answer: b


19. If in ΔABC, ∠C = 90°, then sin (A + B) =
(a) 0
(b) 1/2
(c) \(\frac{1}{\sqrt{2}}\)
(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) \(\frac{3}{4}\)
(d) \(\frac{1}{2}\)

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 = \(\frac{a}{b}\) then the value of

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 \(\frac{2}{3}\)
tan θ = \(\frac{3}{2}\)


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

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 = \(\frac{1}{\cos \theta}\)
∵ 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°

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) \(\frac{1}{\sqrt{3}}\)
(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°) + \(\left(\frac{1}{\sqrt{2}}\right)^{2}\) + 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) \(\frac{3}{2}\)

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 = \(\frac{4}{5}\) ……(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

Answer/Explanation

Answer: c
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

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 = \(\frac{1}{2}\)
⇒ 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 =

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


44. Reciprocal of cot A is _____ .

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 = \(\frac{1}{2}\). Determine the length of side AC.

Answer/Explanation

Answer:
Explaination:


47. If sec θ = \(\frac{25}{7}\), find the values of tan θ and cosec 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
⇒ \(\frac{\sin A}{\cos A}\) = 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 = _____ .

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 = \(\frac{1}{\sqrt{3}}\) =r, then evaluate

Answer/Explanation

Answer:
Explaination:


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

Answer/Explanation

Answer:
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 \(\frac{\tan 65^{\circ}}{\cot 25^{\circ}}\) = _____.

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 = \(\frac{15}{7}\) and A + B = 90°, find the value of cosec B.

Answer/Explanation

Answer:
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 =

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

Answer/Explanation

Answer:c
Explaination:


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

Answer/Explanation

Answer:
Explaination:


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

Answer/Explanation

Answer:
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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