Answer
As left side transforms into right side, hence given identity-
$\frac{1 +\cos x}{1 - \cos x}$ = $(\csc x + \cot x)^{2}$ is true.
Work Step by Step
Given identity is-
$\frac{1 +\cos x}{1 - \cos x}$ = $(\csc x + \cot x)^{2}$
Taking L.S.
$\frac{1 +\cos x}{1 - \cos x}$
= $\frac{1 +\cos x}{1 - \cos x}. \frac{1 +\cos x}{1 + \cos x}$
{Multiplying the numerator and denominator of the fraction by the conjugate of the denominator, $(1 +\cos x)$}
= $\frac{(1 +\cos x)^{2}}{(1 - \cos x)(1 +\cos x)}$
= $\frac{1 + 2\cos x + \cos^{2}x}{{1^{2} - (\cos x)^{2}}}$
{Recall $(a+b)^{2} = a^{2} +2ab +b^{2}$ and
$(a-b)(a+b) $ = $a^{2} - b^{2}$ }
= $\frac{1 + 2\cos x + \cos^{2}x}{1 - \cos ^{2} x}$
= $\frac{1 + 2\cos x + \cos^{2}x}{\sin^{2} x}$
( From first Pythagorean identity)
= $\frac{1}{\sin^{2} x} + \frac{2 \cos x}{\sin^{2} x} + \frac{\cos^{2} x}{\sin^{2} x}$
= $\csc^{2} x + 2.\frac{1}{\sin x}.\frac{\cos x}{\sin x} + \cot^{2} x $
(using ratio and reciprocal identities)
= $\csc^{2} x + 2.\csc x \cot x + \cot^{2} x $
= $(\csc x + \cot x)^{2}$ = R.S.
Hence given identity-
$\frac{1 +\cos x}{1 - \cos x}$ = $(\csc x + \cot x)^{2}$ is true.
