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