Answer
See proof below.
Work Step by Step
Apply the sum and difference formulas:
$\cos(A+B)=\cos(A)\cos(B) -\sin(A)\sin(B)$
and
$\cos(A-B)=\cos(A)\cos(B) +\sin(A) \ \sin(B)$
In order to prove the given identity, we simplify the left hand side $\text{LHS}$ as follows:
$\text{LHS}=\dfrac{\sin{(\alpha+\beta)}}{\sin (\alpha - \beta)}\\
=\dfrac{\sin \alpha \cos \beta+\cos \alpha \sin \beta }{\sin \alpha \cos \beta -\cos \alpha \sin \beta} $
Divide the numerator and denominator by $\cos \alpha \cos \beta$ to obtain:
$\dfrac{\dfrac{\sin \alpha }{\cos \alpha}+\dfrac{\sin \beta }{\cos \beta} }{ \dfrac{\sin \alpha }{\cos \alpha}-\dfrac{\sin \beta }{\cos \beta}} \\=\dfrac{\tan \alpha+\tan \beta}{\tan \alpha -\tan \beta} \\=\text{ RHS}$