Chapter 14, Problem 17MCQ

To determine

To verify: The given identity.

Explanation of Solution

Given:

The identity is given as:

  $\frac{\sin \theta \sec \theta }{\sec \theta -1}=\left(\sec \theta +1\right)\cot \theta$

Consider the identify:

  $\frac{\sin \theta \sec \theta }{\sec \theta -1}=\left(\sec \theta +1\right)\cot \theta$

From identify:

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

Taking the left hand side of the identify:

  $\begin{array}{c}\frac{\sin \theta \sec \theta }{\sec \theta -1}=\frac{\sin \theta \frac{1}{\cos \theta }}{\sec \theta -1}.\frac{\left(\sec \theta +1\right)}{\left(\sec \theta +1\right)}\\ =\frac{\frac{\sin \theta }{\cos \theta }\left(\sec \theta +1\right)}{\left({\sec }^2\theta +1\right)}\end{array}$

From identifies:

  $\begin{array}{l}\tan \theta =\frac{\sin \theta }{\cos \theta }\\ {\sec }^2\theta =1+{\tan }^2\theta \end{array}$

Therefore,

  $\begin{array}{l}\frac{\sin \theta \sec \theta }{\sec \theta -1}=\frac{\tan \theta \left(\sec \theta +1\right)}{{\tan }^2\theta }\\ =\frac{1}{\tan \theta }\left(\sec \theta +1\right)\end{array}$

From identify,

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

Therefore ,

  $\begin{array}{l}\frac{\sin \theta \sec \theta }{\sec \theta -1}=\cot \theta \left(\sec \theta +1\right)\\ =\text{Right}\text{Hand}\text{Side}\end{array}$

Hence, from the above equation, the above identify is verified.