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: 54


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$\tan \,\left( \frac{\pi }{4}+\alpha \right)-\tan \,\left( \frac{\pi }{4}-\alpha \right)$ By using the identities of trigonometry, $\tan \,\left( \alpha +\beta \right)=\frac{\tan \,\alpha +\tan \,\beta }{1-\tan \,\alpha \tan \,\beta }$ And, $\tan \,\left( \alpha -\beta \right)=\frac{\tan \,\alpha -\tan \,\beta }{1+\tan \,\alpha \tan \,\beta }$ Now, the above expression can be further simplified as, $\begin{align} & \tan \,\left( \frac{\pi }{4}+\alpha \right)-\tan \,\left( \frac{\pi }{4}-\alpha \right)=\frac{\tan \,\frac{\pi }{4}+\tan \,\alpha }{1-\tan \,\frac{\pi }{4}\tan \,\alpha }-\frac{\tan \,\frac{\pi }{4}-\tan \,\alpha }{1+\tan \,\frac{\pi }{4}\tan \,\alpha } \\ & =\frac{1+\tan \,\alpha }{1-\tan \,\alpha }-\frac{1-\tan \,\alpha }{1+\tan \,\alpha } \end{align}$ Now, taking LCM of $\left( 1-\tan \alpha \right)$ and $\left( 1+\tan \alpha \right)$ , $\begin{align} & \frac{1+\tan \,\alpha }{1-\tan \,\alpha }-\frac{1-\tan \,\alpha }{1+\tan \,\alpha }=\frac{\left( 1+\tan \,\alpha \right)\left( 1+\tan \,\alpha \right)-\left( 1-\tan \,\alpha \right)\left( 1-\tan \,\alpha \right)}{\left( 1-\tan \,\alpha \right)\left( 1+\tan \,\alpha \right)} \\ & =\frac{\left( 1+\tan \,\alpha +\tan \,\alpha +{{\tan }^{2}}\,\alpha \right)-\left( 1-\tan \,\alpha -\tan \,\alpha +{{\tan }^{2}}\,\alpha \right)}{\left( {{1}^{2}}-{{\tan }^{2}}\,\alpha \right)} \\ & =\frac{\left( 1+2\tan \,\alpha +{{\tan }^{2}}\,\alpha \right)-\left( 1-2\tan \,\alpha +{{\tan }^{2}}\,\alpha \right)}{\left( 1-{{\tan }^{2}}\,\alpha \right)} \\ & =\frac{1+2\tan \,\alpha +{{\tan }^{2}}\,\alpha -1+2\tan \,\alpha -{{\tan }^{2}}\,\alpha }{\left( 1-{{\tan }^{2}}\,\alpha \right)} \end{align}$ The above expression is simplified further, $\begin{align} & \frac{1+2\tan \,\alpha +{{\tan }^{2}}\,\alpha -1+2\tan \,\alpha -{{\tan }^{2}}\,\alpha }{\left( 1-{{\tan }^{2}}\,\alpha \right)}=\frac{2\tan \,\alpha +2\tan \,\alpha }{1-{{\tan }^{2}}\,\alpha } \\ & =\frac{4\tan \,\alpha }{1-{{\tan }^{2}}\,\alpha } \\ & =2\frac{2\tan \,\alpha }{1-{{\tan }^{2}}\,\alpha } \end{align}$ Since, $\tan 2\alpha =\frac{2\tan \alpha }{1-{{\tan }^{2}}\alpha }$. This implies, $2\frac{2\tan \,\alpha }{1-{{\tan }^{2}}\,\alpha }=2\tan \,2\alpha $ Thus, the left side of the given expression is equal to the right side, which is, $\tan \,\left( \frac{\pi }{4}+\alpha \right)-\tan \,\left( \frac{\pi }{4}-\alpha \right)=2\tan \,2\alpha $.
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