## Precalculus (6th Edition) Blitzer

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