Prove the following trigonometric identities.

We have to prove $\frac{\cos ^{2} \theta}{\sin \theta}-\operatorname{cosec} \theta+\sin \theta=0$

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

So,

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

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

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

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

$=-\sin \theta+\sin \theta$

$=0$