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.