Chapter 7.3, Problem 84E

To determine

To prove the identity:

  ${\tan }^2\left(\frac{x}{2}+\frac{\pi }{4}\right)=\frac{1+\sin x}{1-\sin x}$

Explanation of Solution

Given:

An identity is given as

  ${\tan }^2\left(\frac{x}{2}+\frac{\pi }{4}\right)=\frac{1+\sin x}{1-\sin x}$

Concept Used:

The double angle formula for cosine:

  $\sin 2u=2\sin u\cos u$

Pythagorean Identity

  ${\sin }^{2}x+{\cos }^{2}x=1$

Calculation:

In order to verify the identity:

  ${\tan }^2\left(\frac{x}{2}+\frac{\pi }{4}\right)=\frac{1+\sin x}{1-\sin x}$

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

As,

  $\begin{array}{c}{\tan }^2\left(\frac{x}{2}+\frac{\pi }{4}\right)={\left[\tan \left(\frac{x}{2}+\frac{\pi }{4}\right)\right]}^2\\ ={\left[\frac{\tan \frac{x}{2}+\tan \frac{\pi }{4}}{1-\tan \frac{x}{2}\cdot \tan \frac{\pi }{4}}\right]}^2\\ ={\left[\frac{\tan \frac{x}{2}+1}{1-\tan \frac{x}{2}}\right]}^2\\ ={\left[\frac{\sin \frac{x}{2}+\cos \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}\right]}^2\\ =\frac{{\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}}{{\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}-2\sin \frac{x}{2}\cos \frac{x}{2}}\\ =\frac{1+\sin x}{1-\sin x}\\ =\text{Right hand}\end{array}$

Thus, identity is verified, i.e.

  ${\tan }^2\left(\frac{x}{2}+\frac{\pi }{4}\right)=\frac{1+\sin x}{1-\sin x}$