Question

Prove:
$\sqrt{\frac{1+sinA}{1-sinA}}=secA+tanA$

Answer 1

$\sqrt{\frac{1+\sin A}{1-\sin A}}=\sqrt{\frac{(1+\sin A{)}^2}{(1-\sin A)(1+\sin A)}}=\sqrt{\frac{(1+\sin A{)}^2}{{\cos }^{2}A}}=\frac{1+\sin A}{\cos A}=\frac{1}{\cos A}+\frac{\sin A}{\cos A}=\text{sec}A+\tan A$