Chapter 5.2, Problem 52E

To determine

To verify the identity:

  $\frac{1-{\tan }^2\alpha }{1+\cot \left(\alpha -\frac{\pi }{2}\right)}=\frac{\cos \alpha +\sin \alpha }{\cos \alpha }$

And also verify this using table feature of graphing utility.

Explanation of Solution

Given:

  $\frac{1-{\tan }^2\alpha }{1+\cot \left(\alpha -\frac{\pi }{2}\right)}=\frac{\cos \alpha +\sin \alpha }{\cos \alpha }$

Concept Used:

The trigonometric identities:

 • $\cot \left(\frac{\pi }{2}-\alpha \right)=\tan \alpha$

Calculation:

In order to verify the identity:

  $\frac{1-{\tan }^2\alpha }{1+\cot \left(\alpha -\frac{\pi }{2}\right)}=\frac{\cos \alpha +\sin \alpha }{\cos \alpha }$

Take left hand of above identity and simplify it using the trigonometric identities.

As,

  $\begin{array}{c}\frac{1-{\tan }^2\alpha }{1+\cot \left(\alpha -\frac{\pi }{2}\right)}=\frac{\left(1-\tan \alpha \right)\left(1+\tan \alpha \right)}{1-\tan \alpha }\\ =1+\tan \alpha \\ =1+\frac{\sin \alpha }{\cos \alpha }\\ =\frac{\cos \alpha +\sin \alpha }{\cos \alpha }\\ =\text{Right hand}\end{array}$

Thus, identity is verified, i.e.

  $\frac{1-{\tan }^2\alpha }{1+\cot \left(\alpha -\frac{\pi }{2}\right)}=\frac{\cos \alpha +\sin \alpha }{\cos \alpha }$

Table of values for $\frac{1-{\tan }^2\alpha }{1+\cot \left(\alpha -\frac{\pi }{2}\right)}$ is shown:

  

Graph of $\frac{1-{\tan }^2\alpha }{1+\cot \left(\alpha -\frac{\pi }{2}\right)}$ is constructed as

  

Table of values for $\frac{\cos \alpha +\sin \alpha }{\cos \alpha }$ is shown:

  

Graph of $\frac{\cos \alpha +\sin \alpha }{\cos \alpha }$ is constructed as shown:

  

Since, the graph of left hand side and right hand side of the identity coincides, thus the identity is verified by graph too.