Answer
See the full explanation below.
Work Step by Step
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 $\frac{\sin \,\left( \alpha +\beta \right)}{\sin \,\left( \alpha -\beta \right)}=\frac{\tan \,\alpha +\tan \,\beta }{\tan \,\alpha -\tan \,\beta }$.
