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