Question

If $\sec 4A=\csc (A-{20}^o)$, where $4A$ is an acute angle, find the value of $A$.

Answer 1

Consider $\sec 4A$ which can be rewritten as follows:

$\sec 4A=\csc ({90}^0-4A)$

$(\because \csc ({90}^0-x)=\sec x)$

Therefore, $\sec 4A=\csc (A-{20}^0)$ can be evaluated as shown below:

$\sec 4A=\csc (A-{20}^0)\Rightarrow \csc ({90}^0-4A)=\csc (A-{20}^0)\Rightarrow {90}^0-4A=A-{20}^0\Rightarrow {90}^0+{20}^0=A+4A\Rightarrow {110}^0=5A\Rightarrow A=\frac{{110}^0}{5}={22}^0$

Hence, $A={22}^0$.