Problem - 3884
Prove the identity: $\tan^2 x - \sin^2 x = \tan^2 x \sin^2 x$.
\begin{align*}
\tan^2 x - \sin^2 x &=\frac{\sin^2 x}{\cos^2 x}-\sin^2 x\\
&=\Big(\frac{1}{\cos^2 x}-1\Big)\sin^2 x\\
&=(\sec^2 x - 1)\sin^2 x\\
&=\tan^2 x \sin^2 x
\end{align*}