Prove the following trigonometric identities.

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

We know that, $\sin ^{2} \theta+\cos ^{2} \theta=1$

So,

$\cot \theta-\tan \theta=\frac{\cos \theta}{\sin \theta}-\frac{\sin \theta}{\cos \theta}$

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

$=\frac{\cos ^{2} \theta-\left(1-\cos ^{2} \theta\right)}{\sin \theta \cos \theta}$

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

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

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

We know that, $\sin ^{2} \theta+\cos ^{2} \theta=1$

So,

$\tan \theta-\cot \theta=\frac{\sin \theta}{\cos \theta}-\frac{\cos \theta}{\sin \theta}$

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

$=\frac{\sin ^{2} \theta-\left(1-\sin ^{2} \theta\right)}{\sin \theta \cos \theta}$

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

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