## Precalculus (6th Edition) Blitzer

The required value of the $\tan \frac{\alpha }{2}$ is $\pm \sqrt{\frac{1-\cos \,\alpha }{1+\cos \,\alpha }}=\frac{1-\cos \alpha }{\sin \alpha }=\frac{\sin \alpha }{1+\cos \alpha }$.
We have to find the value of $\tan \frac{\alpha }{2}$; the half angle formula is used as follows: \begin{align} & {{\tan }^{2}}\frac{\alpha }{2}=\frac{{{\sin }^{2}}\frac{\alpha }{2}}{{{\cos }^{2}}\frac{\alpha }{2}} \\ & =\frac{\frac{1-\cos 2\frac{\alpha }{2}}{2}}{\frac{1+\cos 2\frac{\alpha }{2}}{2}} \\ & =\frac{1-\cos \alpha }{2}\times \frac{2}{1+\cos \alpha } \\ & =\pm \sqrt{\frac{1-\cos \alpha }{1+\cos \alpha }} \end{align} Then, verify the $\tan \frac{\alpha }{2}$ for the other formula with the half angle formula: $\tan \alpha =\frac{1-\cos 2\alpha }{\sin 2\alpha }$ Replace $\alpha$ with $\frac{\alpha }{2}$ \begin{align} & \tan \frac{\alpha }{2}=\frac{1-\cos 2\frac{\alpha }{2}}{\sin 2\frac{\alpha }{2}} \\ & =\frac{1-\cos \alpha }{\sin \alpha } \end{align} Verify the $\tan \frac{\alpha }{2}$ for the other formula with the half angle formula: $\tan \alpha =\frac{\sin 2\alpha }{1+\cos 2\alpha }$ Replace $\alpha$ with $\frac{\alpha }{2}$ \begin{align} & \tan \frac{\alpha }{2}=\frac{\sin 2\frac{\alpha }{2}}{1+\cos 2\frac{\alpha }{2}} \\ & =\frac{\sin \alpha }{1+\cos \alpha } \end{align} Hence, the value of the $\tan \frac{\alpha }{2}$ is $\pm \sqrt{\frac{1-\cos \,\alpha }{1+\cos \,\alpha }}=\frac{1-\cos \alpha }{\sin \alpha }=\frac{\sin \alpha }{1+\cos \alpha }$.