Answer
See proof below.
Work Step by Step
Apply the sum and difference formulas:
$\sin(A+B)=\sin(A)\cos(B)+\cos(A)\sin(B)$
and
$\sin(A-B)=\sin(A)\cos(B) -\cos(A)\sin(B)$
In order to prove the given identity, we simplify the left hand side $\text{LHS}$ as follows:
$\text{LHS}=\dfrac{\sin{(\alpha+\beta)}}{\sin \alpha \cos \beta}\\
=\dfrac{\sin \alpha \cos \beta+\cos \alpha \sin \beta }{\sin \alpha \cos \beta} \\=\dfrac{\sin \alpha \cos \beta }{\sin \alpha \cos \beta}+\dfrac{\cos \alpha \sin \beta }{\sin \alpha \cos \beta}\\ =1+ \cot \alpha \tan\beta \\=\text{ RHS}$