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: 53


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Work Step by Step

$\tan \,2\alpha $ By using the identity of trigonometry, $\tan \,\left( \alpha +\beta \right)=\frac{\tan \,\alpha +\tan \,\beta }{1-\tan \,\alpha \tan \,\beta }$ Now, the above expression can be further simplified as, $\begin{align} & \tan \,2\alpha =\tan \,\left( \alpha +\alpha \right) \\ & =\frac{\tan \,\alpha +\tan \,\alpha }{1-\tan \,\alpha \tan \,\alpha } \\ & =\frac{2\tan \,\alpha }{1-{{\tan }^{2}}\,\alpha } \end{align}$ Thus, the left side of the given expression is equal to the right side, which is, $\tan \,2\alpha =\frac{2\tan \,\alpha }{1-{{\tan }^{2}}\,\alpha }$.
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