Prove the following trigonometric identities:$ \frac{\left(1+\cot ^{2} \theta\right) \tan \theta}{\sec ^{2} \theta}=\cot \theta $

To do:

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

Solution:

We know that,

$\operatorname{cosec}^2 \theta-cot ^{2} \theta=1$.......(i)

$\cot \theta=\frac{\operatorname{cosec} \theta}{\sec \theta}$........(ii)

$\tan \theta \times \cot \theta=1$........(iii)

Therefore,

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

$=(\frac{\operatorname{cosec} \theta}{\sec \theta})^2\times \tan \theta$

$=(\cot \theta)^2\times \tan \theta$          (From (ii))

$=\cot \theta \times \cot \theta \times \tan \theta$      

$=1\times \cot \theta$           (From (iii))

$=\cot \theta$

Hence proved.