## Precalculus (6th Edition) Blitzer

Let us consider the left side of the given expression: $\frac{\sin \,\left( \alpha +\beta \right)}{\sin \,\left( \alpha -\beta \right)}$ By using the identities of trigonometry, $\sin \,\left( \alpha +\beta \right)=\sin \,\alpha \,\cos \beta +\cos \,\alpha \,\sin \,\beta$ $\sin \,\left( \alpha -\beta \right)=\sin \,\alpha \,\cos \beta -\cos \,\alpha \,\sin \,\beta$ , the above expression can be further simplified as: \begin{align} & \frac{\sin \,\left( \alpha +\beta \right)}{\sin \,\left( \alpha -\beta \right)}=\frac{\sin \,\alpha \,\cos \beta +\cos \,\alpha \,\sin \,\beta }{\sin \,\alpha \,\cos \beta -\cos \,\alpha \,\sin \,\beta } \\ & \text{Divide}\,\text{numerator}\,\text{ and }\,\text{denominator}\,\text{by}\,\cos \,\alpha \cos \,\beta \\ \end{align} \begin{align} & =\frac{\frac{\sin \,\alpha \,\cos \beta }{\cos \,\alpha \cos \,\beta }+\frac{\cos \,\alpha \,\sin \,\beta }{\cos \,\alpha \cos \,\beta }}{\frac{\sin \,\alpha \,\cos \beta }{\cos \,\alpha \cos \,\beta }-\frac{\cos \,\alpha \,\sin \,\beta }{\cos \,\alpha \cos \,\beta }} \\ & =\frac{\frac{\sin \,\alpha }{\cos \,\alpha }+\frac{\sin \,\beta }{\cos \,\beta }}{\frac{\sin \,\alpha }{\cos \,\alpha }-\frac{\sin \,\beta }{\cos \,\beta }} \\ & =\frac{\tan \,\alpha +\tan \,\beta }{\tan \,\alpha -\tan \,\beta } \\ \end{align} Hence, the left side of the given expression is equal to the right side, which is $\frac{\sin \,\left( \alpha +\beta \right)}{\sin \,\left( \alpha -\beta \right)}=\frac{\tan \,\alpha +\tan \,\beta }{\tan \,\alpha -\tan \,\beta }$.