MCQ questions for class 10 Maths Chapter 8- Introduction to Trigonometry are provided here. All these MCQs are provided with answers and are important for CBSE Class 10 Maths Exam which will be conducted on 12th March, 2020. All the questions are also available in PDF. Practice with these questions to revise the important concepts and score good marks in the CBSE Class 10 Maths Exam 2020.

Check below solved MCQs from Class 10 Maths Chapter 8Introduction to Trigonometry:

1. (sin30° + cos30°) – (sin 60° + cos60°)

(A) – 1

(B) 0

(C) 1

(D) 2

Answer: (B)

Explanation: According to question

2. Value of tan30°/cot60° is:

(A) 1/√2

(B) 1/√3

 (C) √3

(D) 1

Answer: (D)

Explanation:

3. sec²θ – 1 = ?

(A) tan²θ

(B) tan²θ + 1

(C) cot²θ – 1

(D) cos²θ

Answer:  (A)

Explanation: From trigonometric identity

1+ tan²θ = sec²θ

⇒sec²θ – 1 = tan²θ

4. The value of sin θ and cos (90° – θ)

(A) Are same

(B) Are different

(C) No relation

(D) Information insufficient

Answer: (A)

Explanation: Since from trigonometric identities,

cos(90° – θ) = sin θ

So, both represents the same value.

5. If cos A = 4/5, then tan A = ?

(A) 3/5

(B) 3/4

(C) 4/3

(D) 4/5

Answer: (B)

Explanation: From trigonometric identity

6. The value of the expression [cosec (75° + θ) – sec (15° - θ) – tan (55° + θ) + cot (35° - θ)] is

(A) 1

(B) −1

(C) 0

(D) 1/2

Answer: (C)

Explanation: Since

cosec (75° + θ) – sec (15° - θ) – tan (55° + θ) + cot (35° - θ)

= cosec (75° + θ) – cosec [90° - (15° - θ)] – tan (55° + θ) + tan [90° - (35° - θ)]

= cosec (75° + θ) – cosec (75° + θ) – tan (55° + θ) + tan (55° + θ)

= 0

7. Given that: SinA = a/b, then cosA = ?

(C) b/a

(D) a/b

Answer:(B)

Explanation: We have

8. The value of (tan1° tan2° tan3° ... tan89°) is

(A) 0

(B) 1

(C) 2

(D)1/2

Answer: (B)

Explanation: This can be written as,

(tan1° tan2° tan3° ... tan89°)

(tan1° tan2° tan3° ....... tan44° tan45° tan46° ..... tan87°tan88°tan89°)

= [tan1° tan2° tan3° ....... tan44° tan45° tan (90 – 44)° ..... tan(90° - 3) tan (90° - 2) tan (90° - 1)]

= (tan1° tan2° tan3° ....... tan44° tan45° cot 44° ..... cot3° cot2° cot 1°)

= 1

Since tan and cot are reciprocals of each other, so they cancel each other.

9. If sin A + sin² A = 1, then cos² A + cos⁴ A = ?

(A) 1

(B) 0

(C) 2

(D) 4

Answer: (A)

Explanation: We have

sin A + sin ² A = 1

⇒ sin A = 1 – sin² A

⇒ sin A = cos² A               ......(i)

Squaring both sides

⇒sin²A = cos⁴A               ......(ii)

From equations (i) and (ii), we have

cos²A + cos⁴A = sin A + sin²A = 1

10. If sin A = 1/2 and cos B = 1/2, then A + B = ?

(A) 0⁰

(B) 30⁰

(C) 60⁰

(D) 90⁰

Answer: (D)

Explanation: Since

(A) 3

(B) 2

(C) 1

(D) 0

Answer: (B)

Explanation: Using trigonometric properties, we have:

12. If cos9α = sin α and 9α 90°, then the value of tan5α is

(A) √3

(B) 1/√3

(C) 0

(D) 1

Answer: (D)

Explanation: Since

cos9α = sinα

⇒ sin (90° - 9α) = sinα

⇒ (90° - 9α) = α

⇒ α = 9°

Therefore,

tan 5α = tan 5 (9°)

= tan45°

= 1

13. If a pole 6m high casts a shadow 2√3 m long on the ground, then the sun’s elevation is

(A) 60°

(B) 45°

(C) 30°

(D)90°

Answer: (A)

Explanation: Given condition can be represented as follows:

14. If cos (A + B) = 0, then sin (A – B) is reduced to:

(A) cos A

(B) cos 2B

(C) sin A

(D) sin 2B

Answer: (B)

Explanation: Since

cos (A + B) = 0

⇒ cos (A + B) = cos90°

⇒ (A + B) = 90°

⇒ A = 90° - B

This implies

sin (A – B) = sin (90° - B - B)

⇒ sin (A – B) = sin (90° - 2B)

sin (A – B) = cos 2B

(A) 2/3

(B) 1/3

(C) 1/2

(D) 3/4

Answer:(C)

Explanation: This can be solved as,