Answer
As left side transforms into right side, hence given identity-
$\frac{\csc x - 1}{\csc x + 1} $ = $ \frac{1 - \sin x}{1 + \sin x}$ is true.
Work Step by Step
Given identity is-
$\frac{\csc x - 1}{\csc x + 1} $ = $ \frac{1 - \sin x}{1 + \sin x}$
Taking L.S.
$\frac{\csc x - 1}{\csc x + 1} $
= $\frac{\frac{1}{\sin x} - 1 }{\frac{1}{\sin x} + 1 } $ (using reciprocal identity)
= $\frac{\frac{1}{\sin x} - \frac{\sin x}{\sin x} } {\frac{1}{\sin x} + \frac{\sin x}{\sin x}} $
= $\frac{\frac{1 -\sin x}{\sin x} }{\frac{1 +\sin x}{\sin x} } $
= $\frac{(1 -\sin x)\sin x} {(1 +\sin x)\sin x} $
= $ \frac{1 - \sin x}{1 + \sin x}$
= R.S.
As left side transforms into right side, hence given identity-
$\frac{\csc x - 1}{\csc x + 1} $ = $ \frac{1 - \sin x}{1 + \sin x}$ is true.