Chapter 5.2, Problem 51E

To determine

To verify the identity:

  $\frac{\sin \alpha \sec \left(\frac{\pi }{2}-\alpha \right)}{1+{\tan }^2\alpha }=\frac{{\cos }^3\alpha }{\sin \left(\frac{\pi }{2}-\alpha \right)}$

And also verify this using table feature of graphing utility.

Explanation of Solution

Given:

  $\frac{\sin \alpha \sec \left(\frac{\pi }{2}-\alpha \right)}{1+{\tan }^2\alpha }=\frac{{\cos }^3\alpha }{\sin \left(\frac{\pi }{2}-\alpha \right)}$

Concept Used:

The trigonometric identities:

 • $\cos \alpha =\sin \left(\frac{\pi }{2}-\alpha \right)$

 • $\sec \left(\frac{\pi }{2}-\alpha \right)=\csc \alpha$

Calculation:

In order to verify the identity:

  $\frac{\sin \alpha \sec \left(\frac{\pi }{2}-\alpha \right)}{1+{\tan }^2\alpha }=\frac{{\cos }^3\alpha }{\sin \left(\frac{\pi }{2}-\alpha \right)}$

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

As,

  $\begin{array}{c}\frac{\sin \alpha \sec \left(\frac{\pi }{2}-\alpha \right)}{1+{\tan }^2\alpha }=\frac{\sin \alpha \sec \left(\frac{\pi }{2}-\alpha \right)}{1+\frac{{\sin }^2\alpha }{{\cos }^2\alpha }}\\ =\frac{\sin \alpha \sec \left(\frac{\pi }{2}-\alpha \right){\cos }^2\alpha }{{\cos }^2\alpha +{\sin }^2\alpha }\\ =\frac{\sin \alpha \csc \alpha {\cos }^2\alpha }{1}\\ =\frac{{\cos }^3\alpha }{\cos \alpha }\\ =\frac{{\cos }^3\alpha }{\sin \left(\frac{\pi }{2}-\alpha \right)}\\ =\text{Right hand}\end{array}$

Thus, identity is verified, i.e.

  $\frac{\sin \alpha \sec \left(\frac{\pi }{2}-\alpha \right)}{1+{\tan }^2\alpha }=\frac{{\cos }^3\alpha }{\sin \left(\frac{\pi }{2}-\alpha \right)}$

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

  

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

  

Table of values for $\frac{{\cos }^3\alpha }{\sin \left(\frac{\pi }{2}-\alpha \right)}$ is shown:

  

Graph of $\frac{{\cos }^3\alpha }{\sin \left(\frac{\pi }{2}-\alpha \right)}$ 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.