Question:
Prove the following trigonometric identities.
$\tan ^{2} \theta-\sin ^{2} \theta=\tan ^{2} \theta \sin ^{2} \theta$
Solution:
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,
$\tan ^{2} \theta-\sin ^{2} \theta=\frac{\sin ^{2} \theta}{\cos ^{2} \theta}-\sin ^{2} \theta$
$=\frac{\sin ^{2} \theta-\sin ^{2} \theta \cos ^{2} \theta}{\cos ^{2} \theta}$
$=\frac{\sin ^{2} \theta\left(1-\cos ^{2} \theta\right)}{\cos ^{2} \theta}$
$=\frac{\sin ^{2} \theta \sin ^{2} \theta}{\cos ^{2} \theta}$
$=\tan ^{2} \theta \sin ^{2} \theta$
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