Prove the following trigonometric identities.

We have to prove $\tan ^{2} \theta-\sin ^{2} \theta=\tan ^{2} \theta \sin ^{2} \theta$

We know that, $\sin ^{2} \theta+\cos ^{2} \theta=1$

So,

$\tan ^{2} \theta-\sin ^{2} \theta=\frac{\sin ^{2} \theta}{\cos ^{2} \theta}-\sin ^{2} \theta$

$=\frac{\sin ^{2} \theta-\sin ^{2} \theta \cos ^{2} \theta}{\cos ^{2} \theta}$

$=\frac{\sin ^{2} \theta\left(1-\cos ^{2} \theta\right)}{\cos ^{2} \theta}$

$=\frac{\sin ^{2} \theta \sin ^{2} \theta}{\cos ^{2} \theta}$

$=\tan ^{2} \theta \sin ^{2} \theta$