Precalculus (6th Edition) Blitzer

The result of $\frac{\sin \left( \alpha +\beta \right)-\sin \left( \alpha -\beta \right)}{\cos \left( \alpha +\beta \right)+\cos \left( \alpha -\beta \right)}$ is $\tan \beta$.
Let us consider the given expression, $\frac{\sin \left( \alpha +\beta \right)-\sin \left( \alpha -\beta \right)}{\cos \left( \alpha +\beta \right)+\cos \left( \alpha -\beta \right)}$ by using the trigonometric identities, $\cos \left( \alpha +\beta \right)=\cos \alpha \cos \beta -\sin \alpha \sin \beta$ , $\cos \left( \alpha -\beta \right)=\cos \alpha \cos \beta +\sin \alpha \sin \beta$ $\sin \left( \alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta$ , $\sin \left( \alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta$ Now, the above expression can be further simplified as: \begin{align} & \frac{\sin \left( \alpha +\beta \right)-\sin \left( \alpha -\beta \right)}{\cos \left( \alpha +\beta \right)+\cos \left( \alpha -\beta \right)}=\frac{\left( \sin \alpha \cos \beta +\cos \alpha \sin \beta \right)-\left( \sin \alpha \cos \beta -\cos \alpha \sin \beta \right)}{\left( \cos \alpha \cos \beta -\sin \alpha \sin \beta \right)+\left( \cos \alpha \cos \beta +\sin \alpha \sin \beta \right)} \\ & =\frac{\sin \alpha \cos \beta +\cos \alpha \sin \beta -\sin \alpha \cos \beta +\cos \alpha \sin \beta }{\cos \alpha \cos \beta -\sin \alpha \sin \beta +\cos \alpha \cos \beta +\sin \alpha \sin \beta } \\ & =\frac{\cos \alpha \sin \beta +\cos \alpha \sin \beta }{\cos \alpha \cos \beta +\cos \alpha \cos \beta } \\ & =\frac{2\cos \alpha \sin \beta }{2\cos \alpha \cos \beta } \end{align} Use, $\frac{\sin x}{\cos x}=\tan x$ \begin{align} & \frac{2\cos \alpha \sin \beta }{2\cos \alpha \cos \beta }=\frac{\sin \beta }{\cos \beta } \\ & =\tan \beta \end{align} Thus, $\frac{\sin \left( \alpha +\beta \right)-\sin \left( \alpha -\beta \right)}{\cos \left( \alpha +\beta \right)+\cos \left( \alpha -\beta \right)}$ can be simplified as $\tan \beta$. Hence, the result of $\frac{\sin \left( \alpha +\beta \right)-\sin \left( \alpha -\beta \right)}{\cos \left( \alpha +\beta \right)+\cos \left( \alpha -\beta \right)}$ is $\tan \beta$.