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