Prove the following trigonometric identities.

We need to prove $\frac{\left(1+\tan ^{2} \theta\right) \cot \theta}{\operatorname{cosec}^{2} \theta}=\tan \theta$

Solving the L.H.S, we get

$\frac{\left(1+\tan ^{2} \theta\right) \cot \theta}{\operatorname{cosec}^{2} \theta}=\frac{\sec ^{2} \theta(\cot \theta)}{\operatorname{cosec}^{2} \theta}$

Using $\sec \theta=\frac{1}{\cos \theta}, \cot \theta=\frac{\cos \theta}{\sin \theta}$ and $\operatorname{cosec} \theta=\frac{1}{\sin \theta}$, we get

$\frac{\sec ^{2} \theta(\cot \theta)}{\operatorname{cosec}^{2} \theta}=\frac{\frac{1}{\cos ^{2} \theta}\left(\frac{\cos \theta}{\sin \theta}\right)}{\frac{1}{\sin ^{2} \theta}}$

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

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

$=\frac{\sin \theta}{\cos \theta}$

$=\tan \theta$

Hence proved.