Question

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

Answer 1

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

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

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

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

$=\sec A+\tan A$

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