Prove the following trigonometric identities.

We know that, $\sec ^{2} \theta-\tan ^{2} \theta=1$

So,

$\tan \theta+\frac{1}{\tan \theta}=\frac{\tan ^{2} \theta+1}{\tan \theta}$

$=\frac{\sec ^{2} \theta}{\tan \theta}$

$=\sec \theta \frac{\sec \theta}{\tan \theta}$

$=\sec \theta \frac{\frac{1}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}}$

$=\sec \theta \frac{1}{\sin \theta}$

$=\sec \theta \operatorname{cosec} \theta$