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