We start with the left-hand side (LHS):

\(\tan^2 x - \sin^2 x = \frac{\sin^2 x}{\cos^2 x} - \sin^2 x\)

\(= \frac{\sin^2 x - \sin^2 x \cos^2 x}{\cos^2 x}\)

\(= \frac{\sin^2 x (1 - \cos^2 x)}{\cos^2 x}\)

Using the identity \(1 - \cos^2 x = \sin^2 x\), we have:

\(= \frac{\sin^2 x \sin^2 x}{\cos^2 x}\)

\(= \tan^2 x \sin^2 x\)

Thus, the identity \(\tan^2 x - \sin^2 x \equiv \tan^2 x \sin^2 x\) is proven.