Prove the following identity,where the angle involved is an acute angle for which the expression is defined:
$\sqrt{\frac{1+\sin A}{1-\sin A}} = \sec A + \tan A$

(N/A) To prove the identity $\sqrt{\frac{1+\sin A}{1-\sin A}} = \sec A + \tan A$:
$L.H.S. = \sqrt{\frac{1+\sin A}{1-\sin A}}$
Multiply the numerator and denominator inside the square root by $(1 + \sin A)$:
$= \sqrt{\frac{(1+\sin A)(1+\sin A)}{(1-\sin A)(1+\sin A)}}$
Using the identity $(1 - \sin^2 A) = \cos^2 A$:
$= \sqrt{\frac{(1+\sin A)^2}{1-\sin^2 A}} = \sqrt{\frac{(1+\sin A)^2}{\cos^2 A}}$
Taking the square root:
$= \frac{1+\sin A}{\cos A}$
$= \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \sec A + \tan A$
$= R.H.S.$