Prove the following trigonometric identities:$ \cot \theta-\tan \theta=\frac{2 \cos ^{2} \theta-1}{\sin \theta \cos \theta} $

To do:

We have to prove that \( \cot \theta-\tan \theta=\frac{2 \cos ^{2} \theta-1}{\sin \theta \cos \theta} \).

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,

$\cot \theta-\tan \theta=\frac{\cos \theta}{\sin \theta}-\frac{\sin \theta}{\cos \theta}$               [From (i) and (ii)]

$=\frac{\cos^2 \theta-\sin^2 \theta}{\sin \theta\cos \theta}$          

$=\frac{\cos^2 \theta-(1-\cos^2 \theta)}{\sin \theta\cos \theta}$                    [From (iii)] 

$=\frac{2\cos^2 \theta-1}{\sin \theta\cos \theta}$              

Hence proved.