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