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