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