Question

Evaluate: $\sqrt{\frac{1+\sin A}{1-\sin A}}=\sec A+\tan A$

Answer 1

$\sqrt{\frac{1+\sin A}{1-\sin A}}=\sec A+\tan A$

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

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

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

$=\sec A+\tan A$.