## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 5 - Section 5.2 - Sum and Difference Formulas - Exercise Set - Page 669: 9

#### Answer

See the full explanation below.

#### Work Step by Step

Evaluate the term $\cos \left( \alpha -\beta \right)$ using the cosines difference formula and solve the expression on the left-hand side of the identity as, \begin{align} & \frac{\cos \left( \alpha -\beta \right)}{\cos \alpha \sin \beta }=\frac{\cos \alpha \cos \beta +\sin \alpha \sin \beta }{\cos \alpha \sin \beta } \\ & =\frac{\cos \alpha \cos \beta }{\cos \alpha \sin \beta }+\frac{\sin \alpha \sin \beta }{\cos \alpha \sin \beta } \\ & =\frac{\cos \beta }{\sin \beta }+\frac{\sin \alpha }{\cos \alpha } \end{align} Substitute $\frac{\sin \alpha }{\cos \alpha }=\tan \alpha \text{ and }\frac{\cos \beta }{\sin \beta }=\cot \beta$. $\frac{\cos \left( \alpha -\beta \right)}{\cos \alpha \sin \beta }=\tan \beta +\tan \alpha$ Since the left-hand side part of the identity is equivalent to the expression on the right-hand side, therefore, the identity is verified.

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