Trigonometry — Ratios & Identities — Practice

Question 1

If tan θ = 8/15 (θ acute), then cosec θ equals:
(a) 17/8  (b) 17/15  (c) 15/17  (d) 8/17

Accepted Answer

8-15-17 triple: P = 8, B = 15, H = 17.

cosec θ = H/P = 17/8.

Answer: (a) 17/8.

Question 2

The value of sin²30° + cos²45° + tan²60° is:
(a) 3  (b) 15/4  (c) 7/2  (d) 13/4

Accepted Answer

= (1/2)² + (1/√2)² + (√3)² = 1/4 + 1/2 + 3 = 15/4.

Answer: (b) 15/4.

Question 3

If sec θ − tan θ = 1/3, then sec θ + tan θ equals:
(a) 1/3  (b) 2/3  (c) 3  (d) 9

Accepted Answer

(sec−tan)(sec+tan) = sec²−tan² = 1.

So sec θ + tan θ = 1 ÷ (1/3) = 3.

Answer: (c) 3.

Question 4

If sec θ + tan θ = 2, then cos θ equals:
(a) 3/5  (b) 4/5  (c) 5/4  (d) 5/3

Accepted Answer

sec − tan = 1/2 → sec θ = (2 + 1/2)/2 = 5/4.

cos θ = 4/5.

Answer: (b) 4/5.

Question 5

The value of tan 10° · tan 20° · tan 45° · tan 70° · tan 80° is:
(a) 0  (b) 1  (c) √3  (d) 1/√3

Accepted Answer

tan10°·tan80° = 1 and tan20°·tan70° = 1 (complementary pairs); tan45° = 1.

Product = 1.

Answer: (b) 1.

Question 6

The maximum value of 7 sin θ + 24 cos θ is:
(a) 24  (b) 25  (c) 31  (d) 17

Accepted Answer

Max = √(7² + 24²) = √625 = 25 (7-24-25 triple).

Answer: (b) 25. Trap: 31 = 7 + 24.

Question 7

The minimum value of sin⁴θ + cos⁴θ is:
(a) 0  (b) 1/4  (c) 1/2  (d) 1

Accepted Answer

sin⁴+cos⁴ = 1 − 2sin²θcos²θ = 1 − (1/2)sin²2θ.

Minimum when sin²2θ = 1 (θ = 45°): 1 − 1/2 = 1/2.

Answer: (c) 1/2.

Question 8

If cos θ + cos²θ = 1, then sin²θ + sin⁴θ equals:
(a) 0  (b) 1/2  (c) 1  (d) 2

Accepted Answer

From the given: cos θ = 1 − cos²θ = sin²θ.

Then sin²θ + sin⁴θ = cos θ + cos²θ = 1.

Answer: (c) 1. (The mirror twin of the sinθ + sin²θ = 1 classic.)

Question 9

Simplify: (1 + tan²θ)(1 − sin θ)(1 + sin θ)
(a) 0  (b) 1  (c) sin²θ  (d) cos²θ

Accepted Answer

(1 − sinθ)(1 + sinθ) = 1 − sin²θ = cos²θ; (1 + tan²θ) = sec²θ.

sec²θ × cos²θ = 1.

Answer: (b) 1.

Question 10

The value of sin²5° + sin²85° + sin²45° is:
(a) 1  (b) 3/2  (c) 2  (d) 1/2

Accepted Answer

sin85° = cos5°, so sin²5° + sin²85° = sin²5° + cos²5° = 1.

Add sin²45° = 1/2 → total 3/2.

Answer: (b) 3/2.

Question 11

The value of tan 75° + cot 75° is:
(a) 2√3  (b) 4  (c) 2 + √3  (d) 2 − √3

Accepted Answer

tan75° = 2 + √3 and cot75° = tan15° = 2 − √3.

Sum = 4.

Answer: (b) 4. (Or: tanθ + cotθ = 1/(sinθcosθ) = 2/sin2θ = 2/sin150° = 2/(1/2) = 4.)

Practice — Trigonometry — Ratios & Identities

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