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