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