Question

Prove that :
$\sqrt{\frac{1+\sin A}{1-\sin A}}=(\sec A+\tan A)$

Answer 1

$\sqrt{\frac{1+\sin A}{1-\sin A}}$

$=\sqrt{{\left(\frac{1+\sin A}{\cos A}\right)}^2}$

$=\frac{1+\sin A}{\cos A}$

$=\frac{1}{\cos A}+\frac{\sin A}{\cos A}$

$=\sec A+\tan A$