Answer
As left side transforms into right side, hence given identity-
$\frac{1}{1 - \sin x} + \frac{1}{1 + \sin x}$ = $2 \sec^{2} x$ is true.
Work Step by Step
Given identity is-
$\frac{1}{1 - \sin x} + \frac{1}{1 + \sin x}$ = $2 \sec^{2} x$
Taking L.S.
$\frac{1}{1 - \sin x} + \frac{1}{1 + \sin x}$
= $\frac{(1 + \sin x) + (1 - \sin x) }{(1 - \sin x) (1 +\sin x)} $
{(1 - \sin x) (1 +\sin x) is the L.C.M.}
= $\frac{(1 + \sin x + 1 - \sin x) }{(1 - \sin x) (1 +\sin x)} $
= $\frac{2 }{(1 - \sin x) (1 +\sin x)} $
= $\frac{2 }{1 - \sin^{2} x} $
{Recall $(a-b)(a+b) $ = $a^{2} - b^{2}$ }
= $\frac{2 }{ \cos^{2} x} $
( From first Pythagorean identity, $1 - \sin^{2} x$ = $\cos^{2} x$)
= 2 $\frac{1 }{\cos^{2} x} $
= $2 \sec^{2} x$
= R.S.
As left side transforms into right side, hence given identity-
$\frac{1}{1 - \sin x} + \frac{1}{1 + \sin x}$ = $2 \sec^{2} x$ is true.
