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 tan 5α 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,

Important Link

Quick Revision Notes : Introduction to Trigonometry

NCERT Solution : Introduction to Trigonometry

MCQs: Introduction to Trigonometry

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