Question:
If $\sec 2 A=\operatorname{cosec}\left(A-42^{\circ}\right)$, where $2 A$ is an acute angles, find the value of $A$.
Solution:
Given: $\sec 2 A=\operatorname{cosec}\left(A-42^{\circ}\right)$ and $2 A$ is an acute angle
We have to find $\theta$
So we proceed as follows to calculate $\theta$
$\sec 2 A=\operatorname{cosec}\left(A-42^{\circ}\right)$
$\Rightarrow \sec 2 A=\sec \left\{90^{\circ}-\left(A-42^{\circ}\right)\right\}$
$\Rightarrow \sec 2 A=\sec \left(90^{\circ}-A+42^{\circ}\right)$
$\Rightarrow \sec 2 A=\sec \left(132^{\circ}-A\right)$
$\Rightarrow 3 A=132^{\circ}$
$\Rightarrow A=44^{\circ}$
Hence the value of $A$ is $44^{\circ}$