Answer
As left side transforms into right side, hence given identity-
$ \frac{\sin^{2} B - \tan^{2} B}{1 - \sec^{2} B}$ = $\sin^{2} B $ is true.
Work Step by Step
Given identity is-
$ \frac{\sin^{2} B - \tan^{2} B}{1 - \sec^{2} B}$ = $\sin^{2} B $
Taking L.S.
$ \frac{\sin^{2} B - \tan^{2} B}{1 - \sec^{2} B}$
= $ \frac{-1(\tan^{2} B - \sin^{2} B )}{-1(\sec^{2} B - 1)}$
(taking (-1) out as common factor to change sign to suite trigonometric identities)
= $ \frac{\tan^{2} B - \sin^{2} B }{\sec^{2} B - 1}$
{(-1) cancelled out as common factor}
= $ \frac{\tan^{2} B - \sin^{2} B}{\tan^{2} B}$
( From second Pythagorean identity, $\sec^{2} B - 1= \tan^{2} B$)
= $ \frac{\tan^{2} B }{\tan^{2} B} - \frac{\sin^{2} B}{\tan^{2} B}$
= $ 1- \sin^{2} B \frac{1}{\tan^{2} B}$
= $ 1- \sin^{2} B \cot^{2} B$
= $ 1- \sin^{2} B \frac{\cos^{2} B}{\sin^{2} B}$
= $ 1- \cos^{2} B $
= $\sin^{2} B $ = R.S.
( From first Pythagorean identity, $1 -\cos^{2}B$ can be replaced with, $\sin^{2}B$)
As left side transforms into right side, hence given identity-
$ \frac{\sin^{2} B - \tan^{2} B}{1 - \sec^{2} B}$ = $\sin^{2} B $ is true.