Answer
As left side transforms into right side, hence given identity-
$ \frac{\cot^{2} B - \cos^{2} B}{\csc^{2} B - 1}$ = $\cos^{2} B $ is true.
Work Step by Step
Given identity is-
$ \frac{\cot^{2} B - \cos^{2} B}{\csc^{2} B - 1}$ = $\cos^{2} B $
Taking L.S.
$ \frac{\cot^{2} B - \cos^{2} B}{\csc^{2} B - 1}$
= $ \frac{\cot^{2} B - \cos^{2} B}{\cot^{2} B}$
( From third Pythagorean identity, $\csc^{2} B - 1= \cot^{2} B$)
= $ \frac{\cot^{2} B }{\cot^{2} B}$ - $ \frac{\cos^{2} B}{\cot^{2} B}$
= $ 1- \cos^{2} B \frac{1}{\cot^{2} B}$
= $ 1- \cos^{2} B \tan^{2} B$
= $ 1- \cos^{2} B \frac{\sin^{2} B}{\cos^{2} B}$
= $ 1- \sin^{2} B $
= $\cos^{2} B $ = R.S.
( From first Pythagorean identity, $1 -\sin^{2}B$ can be replaced with, $\cos^{2}B$)
As left side transforms into right side, hence given identity-
$ \frac{\cot^{2} B - \cos^{2} B}{\csc^{2} B - 1}$ = $\cos^{2} B $ is true.
