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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Updated on: 10-Oct-2022

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