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:
$\text{LHS}=\dfrac{\cos{(\alpha -\beta)}}{\sin \alpha \cos \beta}\\
=\dfrac{\cos \alpha \cos \beta +\sin \alpha \sin \beta }{\sin \alpha \cos \beta} \\=\dfrac{\cos \alpha \cos \beta }{\sin \alpha \cos \beta} + \dfrac{\sin \alpha \sin \beta }{\sin \alpha \cos \beta}\\ =\cot \alpha + \ \tan\beta \\=\text{ RHS}$