Chapter 5.2, Problem 65E

To determine

To verify the identity:

  $\sqrt{\frac{1+\sin \alpha }{1-\sin \alpha }=}=\frac{1+\sin \alpha }{\left|\cos \alpha \right|}$

And also verify this using graphing utility.

Explanation of Solution

Given:

  $\sqrt{\frac{1+\sin \alpha }{1-\sin \alpha }=}=\frac{1+\sin \alpha }{\left|\cos \alpha \right|}$

Concept Used:

The trigonometric identity:

  ${\sin }^2\theta +{\cos }^2\theta =1$

Calculation:

In order to verify the identity:

  $\sqrt{\frac{1+\sin \alpha }{1-\sin \alpha }=}=\frac{1+\sin \alpha }{\left|\cos \alpha \right|}$

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

As,

  $\begin{array}{c}\sqrt{\frac{1+\sin \alpha }{1-\sin \alpha }}=\frac{\sqrt{1+\sin \alpha }}{\sqrt{1-\sin \alpha }}\\ \text{= }\frac{\sqrt{1+\sin \alpha }}{\sqrt{1-\sin \alpha }}\times \frac{\sqrt{1+\sin \alpha }}{\sqrt{1+\sin \alpha }},\text{by multiplying numerator and denominator by }\sqrt{1+\sin \alpha }\\ =\frac{{\left(\sqrt{1+\sin \alpha }\right)}^2}{\left(\sqrt{1-\sin \alpha }\sqrt{1+\sin \alpha }\right)}\\ =\frac{1+\sin \alpha }{\sqrt{\left(1-\sin \alpha \right)\left(1+\sin \alpha \right)}}\\ =\frac{1+\sin \alpha }{\sqrt{1^2-{\sin }^2\alpha }}\\ =\frac{1+\sin \alpha }{\sqrt{1-{\sin }^2\alpha }}\\ =\frac{1+\sin \alpha }{\sqrt{{\cos }^2\alpha }},\text{ using identity }\left({\sin }^2\theta +{\cos }^2\theta =1\right)\\ =\frac{1+\sin \alpha }{\left|\cos \alpha \right|}\\ =\text{Right Hand}\end{array}$

Thus, identity is verified, i.e.

  $\sqrt{\frac{1+\sin \alpha }{1-\sin \alpha }=}=\frac{1+\sin \alpha }{\left|\cos \alpha \right|}$

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

  

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