Answer
As left side transforms into right side, hence given identity-
$\frac{\cos x}{1 + \sin x} - \frac{1 - \sin x}{\cos x}$ = $0$ is true.
Work Step by Step
Given identity is-
$\frac{\cos x}{1 + \sin x} - \frac{1 - \sin x}{\cos x}$ = $0$
Taking L.S.
$\frac{\cos x}{1 + \sin x} - \frac{1 - \sin x}{\cos x}$
= $\frac{\cos^{2} x - (1 - \sin x)(1 + \sin x) }{(1 + \sin x) \cos x} $
= $\frac{\cos^{2} x - (1^{2} - \sin^{2} x ) }{(1 + \sin x) \cos x} $
{Recall $(a-b)(a+b) $ = $a^{2} - b^{2}$ }
= $\frac{\cos^{2} x - (1 - \sin^{2} x ) }{(1 + \sin x) \cos x} $
= $\frac{\cos^{2} x - (\cos^{2} x ) }{(1 + \sin x) \cos x} $
( Recall first Pythagorean identity, $1 - \cos^{2} x =\sin^{2} x$ )
= $\frac{\cos^{2} x - \cos^{2} x }{(1 + \sin x) \cos x} $
= $\frac{0 }{(1 + \sin x) \cos x} $
= $0$
= R.S.
As left side transforms into right side, hence given identity-
$\frac{\cos x}{1 + \sin x} - \frac{1 - \sin x}{\cos x}$ = $0$ is true.