Question

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

Answer 1

Given: sec4A = cosec(A – 15°)

$$\begin{gathered} \sec 4A=\mathrm{cosec}\left(A-15°\right) \\ \Rightarrow \mathrm{cosec}\left(90°-4A\right)=\mathrm{cosec}\left(A-15°\right)\left(\because \sec \theta =\mathrm{cosec}\left(90°-\theta \right)\right) \\ \Rightarrow 90°-4A=A-15° \\ \Rightarrow 90°+15°=A+4A \\ \Rightarrow 5A=105° \\ \Rightarrow A=\frac{105°}{5} \\ \Rightarrow A=21° \\ \mathrm{Hence},\angle A=21°. \end{gathered}$$