## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 5 - Review Exercises - Page 708: 23

#### 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 $\frac{\cos \left( \alpha -\beta \right)}{\cos \alpha \cos \beta }$, which can be simplified by using the difference formula $\cos \left( x-y \right)$ that is equal to the product of cosine of the first angle and the cosine of the second angle plus the product of the sine of the first angle and the second angle. $\cos \left( x-y \right)=\cos x\cos y+\sin x\sin y$ We use the above formula along with the quotient identity $\frac{\sin x}{\cos x}=\tan x$ in the expression for further simplification. \begin{align} & \frac{\cos \left( \alpha -\beta \right)}{\cos \alpha \cos \beta }=\frac{\cos \alpha \cos \beta +\sin \alpha \sin \beta }{\cos \alpha \cos \beta } \\ & =\frac{\cos \alpha \cos \beta }{\cos \alpha \cos \beta }+\frac{\sin \alpha \sin \beta }{\cos \alpha \cos \beta } \\ & =1+\tan \alpha \tan \beta \end{align} Hence, the expression on the left-side is equal to the expression on the right-side.

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