## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 5 - Review Exercises - Page 708: 21

#### Answer

The expression on the left-side is equal to the expression on the right-side.

#### Work Step by Step

Let us consider the given expression on the left side $\tan \left( x+\frac{3\pi }{4} \right)$. Now, it can be simplified by using the identity: $\tan \left( \theta +\phi \right)=\frac{\tan \theta +\tan \phi }{1-\tan \theta \tan \phi }$ \begin{align} & \tan \left( x+\frac{3\pi }{4} \right)=\frac{\tan x+\tan \frac{3\pi }{4}}{1-\tan x.\tan \frac{3\pi }{4}} \\ & =\frac{\tan x+\left( -1 \right)}{1-\tan x.\left( -1 \right)} \\ & =\frac{\tan x-1}{1+\tan x} \end{align} Therefore, the expression can be further simplified, where $\theta =x$ and $\phi =\frac{3\pi }{4}$. Hence, the given expression on the left side is equal to the expression on the right side. $\tan \left( x+\frac{3\pi }{4} \right)=\frac{\tan x-1}{1+\tan x}$.

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