Answer
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 $.
Work Step by Step
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 $.
