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