Answer
As left side transforms into right side, hence given identity-
$\csc^{4} \theta - \cot^{4} \theta$ = $\frac{1 + \cos^{2}\theta}{\sin^{2}\theta}$ is true.
Work Step by Step
Given identity is-
$\csc^{4} \theta - \cot^{4} \theta$ = $\frac{1 + \cos^{2}\theta}{\sin^{2}\theta}$
Taking L.S.
$\csc^{4} \theta - \cot^{4} \theta$
= $(\csc^{2} \theta)^{2} - (\cot^{2} \theta)^{2}$
= $(\csc^{2} \theta - \cot^{2} \theta)$ $(\csc^{2} \theta + \cot^{2} \theta)$
{Recall $a^{2} - b^{2}$ = (a-b)(a+b)}
= $(\csc^{2} \theta + \cot^{2} \theta)$
( From second Pythagorean identity, $(\csc^{2} \theta - \cot^{2} \theta)$ = 1)
= $\frac{1}{\sin^{2} \theta} + \frac{\cos^{2} \theta}{\sin^{2} \theta}$
( Using reciprocal identities)
= $\frac{1 + \cos^{2}\theta}{\sin^{2}\theta}$
= R.S.
As left side transforms into right side, hence given identity-
$\csc^{4} \theta - \cot^{4} \theta$ = $\frac{1 + \cos^{2}\theta}{\sin^{2}\theta}$ is true.
