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Prove the following trigonometric identities:$ \tan ^{2} \theta-\sin ^{2} \theta=\tan ^{2} \theta \sin ^{2} \theta $
To do:
We have to prove that \( \tan ^{2} \theta-\sin ^{2} \theta=\tan ^{2} \theta \sin ^{2} \theta \).
Solution:
We know that,
$\tan \theta=\frac{\sin \theta}{\cos \theta}$........(i)
$\sin^2 \theta+cos ^{2} \theta=1$.......(ii)
Therefore,
$\tan ^{2} \theta-\sin ^{2} \theta=\frac{\sin ^{2} \theta}{\cos^2 \theta}-\sin ^{2} \theta$ (From (i))
$=\frac{\sin ^{2} \theta-\sin ^{2} \theta\cos ^{2} \theta}{\cos^2 \theta}$
$=\frac{\sin ^{2} \theta(1-\cos ^{2} \theta)}{\cos^2 \theta}$
$=\sin ^{2} \theta (\frac{\sin ^{2} \theta}{\cos^2 \theta})$ (From (ii))
$=\sin ^{2} \theta \times \tan ^{2} \theta$
$=\tan ^{2} \theta\sin ^{2} \theta$
Hence proved.
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