Question

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

Answer 1

$=\sqrt{\left[\frac{{(1-\sin A)}^2}{(1-{\sin }^{2}A)}\right]}=\frac{1-\sin A}{\cos A}=\frac{1}{\cos A}-\frac{\sin A}{\cos A}=\sec A-\tan A$.