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

Given: sec4A = cosec(A – 15°)

$\sec 4 A=\operatorname{cosec}\left(A-15^{\circ}\right)$

$\Rightarrow \operatorname{cosec}\left(90^{\circ}-4 A\right)=\operatorname{cosec}\left(A-15^{\circ}\right) \quad\left(\because \sec \theta=\operatorname{cosec}\left(90^{\circ}-\theta\right)\right)$

$\Rightarrow 90^{\circ}-4 A=A-15^{\circ}$

$\Rightarrow 90^{\circ}+15^{\circ}=A+4 A$

$\Rightarrow 5 A=105^{\circ}$

$\Rightarrow A=\frac{105^{\circ}}{5}$

$\Rightarrow A=21^{\circ}$

Hence, $\angle A=21^{\circ}$.