Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

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


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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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