Chapter 5, Problem 9CT

To determine

To verify the identity:

  $\frac{\csc \alpha +\sec \alpha }{\sin \alpha +\cos \alpha }=\tan \alpha +\cot \alpha$

And also verify this using graphing utility.

Explanation of Solution

Given:

  $\frac{\csc \alpha +\sec \alpha }{\sin \alpha +\cos \alpha }=\tan \alpha +\cot \alpha$

Concept Used:

The trigonometric identities:

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

Calculation:

In order to verify the identity:

  $\frac{\csc \alpha +\sec \alpha }{\sin \alpha +\cos \alpha }=\tan \alpha +\cot \alpha$

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

As,

  $\begin{array}{c}\frac{\csc \alpha +\sec \alpha }{\sin \alpha +\cos \alpha }=\frac{\frac{1}{\sin \alpha }+\frac{1}{\cos \alpha }}{\sin \alpha +\cos \alpha }\\ =\frac{\frac{\sin \alpha +\cos \alpha }{\sin \alpha \cos \alpha }}{\sin \alpha +\cos \alpha }\\ =\frac{1}{\sin \alpha \cos \alpha }\\ =\frac{{\sin }^2\alpha +{\cos }^2\alpha }{\sin \alpha \cos \alpha }\\ =\frac{{\sin }^2\alpha }{\sin \alpha \cos \alpha }+\frac{{\cos }^2\alpha }{\sin \alpha \cos \alpha }\\ =\frac{\sin \alpha }{\cos \alpha }+\frac{\cos \alpha }{\sin \alpha }\\ =\tan \alpha +\cot \alpha \\ =\text{Right Hand}\end{array}$

Thus, identity is verified, i.e.

  $\frac{\csc \alpha +\sec \alpha }{\sin \alpha +\cos \alpha }=\tan \alpha +\cot \alpha$

Now to verify this identity draw graph of the left side and right side of this side and whether the both of them coincides.

  

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