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: 42


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Work Step by Step

Let us consider the left side of the given expression: $\frac{\sin \left( \alpha +\beta \right)}{\cos \,\alpha \cos \,\beta }$ By using the identity of trigonometry, $\sin \,\left( \alpha +\beta \right)=\sin \,\alpha \,\cos \beta +\cos \,\alpha \,\sin \,\beta $ and $\frac{\sin \,\alpha }{\cos \,\alpha }=\tan \,\alpha $ , the above expression can be further simplified as: $\begin{align} & \frac{\sin \left( \alpha +\beta \right)}{\cos \,\alpha \cos \,\beta }=\frac{\sin \,\alpha \,\cos \beta +\cos \,\alpha \,\sin \,\beta }{\cos \,\alpha \cos \,\beta } \\ & =\frac{\sin \,\alpha \,\cos \beta }{\cos \,\alpha \cos \,\beta }+\frac{\cos \,\alpha \,\sin \beta }{\cos \,\alpha \cos \,\beta } \\ & =\frac{\sin \,\alpha }{\cos \,\alpha }+\frac{\sin \beta }{\cos \,\beta } \\ & =\tan \,\alpha +\tan \,\beta \end{align}$ Hence, the left side of the given expression is equal to the right side, which is $\frac{\sin \left( \alpha +\beta \right)}{\cos \,\alpha \cos \,\beta }=\tan \,\alpha +\tan \,\beta $.
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