Prove the following trigonometric identities.

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

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

So,

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

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

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

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

$=\frac{1}{\tan \theta}$

$=\cot \theta$