## Precalculus (6th Edition) Blitzer

$\tan \,2\alpha$ 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, $\tan \,2\alpha =\frac{2\tan \,\alpha }{1-{{\tan }^{2}}\,\alpha }$.