## Precalculus (6th Edition) Blitzer

Let us consider the right side of the given expression: $\frac{{{\sin }^{2}}\alpha +1-{{\cos }^{2}}\alpha }{\sin \alpha \left( 1+\cos \alpha \right)}$ By using the trigonometric identity $\tan \frac{x}{2}=\frac{\sin x}{1+\cos x}$ and ${{\sin }^{2}}x=1-{{\cos }^{2}}x$, The above expression can be further simplified as: \begin{align} & \frac{{{\sin }^{2}}\alpha +1-{{\cos }^{2}}\alpha }{\sin \alpha \left( 1+\cos \alpha \right)}=\frac{{{\sin }^{2}}\alpha +{{\sin }^{2}}\alpha }{\sin \alpha \left( 1+\cos \alpha \right)} \\ & =\frac{2{{\sin }^{2}}\alpha }{\sin \alpha \left( 1+\cos \alpha \right)} \\ & =2\left( \frac{\sin \alpha }{1+\cos \alpha } \right) \\ & =2\tan \frac{\alpha }{2} \end{align} Hence, the left side of the given expression is equal to the right side, which is $2\tan \frac{\alpha }{2}=\frac{{{\sin }^{2}}\alpha +1-{{\cos }^{2}}\alpha }{\sin \alpha \left( 1+\cos \alpha \right)}$.