Question

If sec 4A = cosec (A – 15°), where 4A is acute then find ∠A.

Answer 1

Given: sec 4A = cosec(A – 15°).

Step-wise calculation:

1. Write in sine/cosine:

`1/(cos4A) = 1/(sin(A - 15^circ)` 

⇒ sin(A – 15°) = cos 4A

2. Use cos θ = sin(90° – θ):

sin(A – 15°) = sin(90° – 4A)

3. Therefore (for integer k) either

(i) A – 15° = 90° – 4A + 360k

⇒ 5A = 105° + 360k

⇒ A = 21° + 72k 

(ii) A – 15° = 180° – (90° – 4A) + 360k

= 90° + 4A + 360k

⇒ –3A = 105° + 360k 

⇒ A = –35° – 120k

4. Condition: 4A is acute

 ⇒ 0° < 4A < 90°

⇒ 0° < A < 22.5°

From the general solutions, the only value in (0, 22.5) is A = 21° take k = 0 in (i). The family in (ii) gives no valid A in this interval.

∠A = 21°.