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)}}{\cos (\alpha - \beta)}\\
=\dfrac{\cos \alpha \cos \beta-\sin \alpha \sin \beta }{\cos \alpha \cos \beta +\sin \alpha \sin \beta} $
Divide the numerator and denominator by $\cos \alpha \cos \beta$ to obtain:
$\dfrac{1-\dfrac{\sin \alpha }{\cos \alpha}\dfrac{\sin \beta }{\cos \beta} }{ 1 +\dfrac{\sin \alpha }{\cos \alpha}\dfrac{\sin \beta }{\cos \beta}} \\=\dfrac{1-\tan \alpha \ \tan \beta}{1 + \tan \alpha \ \tan \beta} \\=\text{ RHS}$