Answer
See the full explanation below.
Work Step by Step
Let us consider the left side of the given expression:
$\frac{\cos \left( \alpha +\beta \right)}{\cos \left( \alpha -\beta \right)}$
By using the identities of trigonometry,
$\cos \left( \alpha +\beta \right)=\cos \alpha \cos \beta -\sin \alpha \sin \beta $
$\cos \left( \alpha -\beta \right)=\cos \alpha \cos \beta +sin\alpha \sin \beta $ ,
Now above provided expression can be further simplified as:
$\frac{\cos \left( \alpha +\beta \right)}{\cos \left( \alpha -\beta \right)}=\frac{\cos \alpha \cos \beta -\sin \alpha \sin \beta }{\cos \alpha \cos \beta +sin\alpha \sin \beta }$
By dividing the numerator and denominator by the $\cos \,\alpha \cos \,\beta $, we get:
$\begin{align}
& \frac{\cos \alpha \cos \beta -\sin \alpha \sin \beta }{\cos \alpha \cos \beta +sin\alpha \sin \beta }=\frac{\frac{\cos \alpha \cos \beta }{\cos \alpha \cos \beta }-\frac{\sin \alpha \sin \beta }{\cos \alpha \cos \beta }}{\frac{\cos \alpha \cos \beta }{\cos \alpha \cos \beta }+\frac{\sin \alpha \sin \beta }{\cos \alpha \cos \beta }} \\
& =\frac{1-\frac{\sin \alpha }{\cos \alpha }\frac{\sin \beta }{\cos \beta }}{1+\frac{\sin \alpha }{\cos \alpha }\frac{\sin \beta }{\cos \beta }} \\
& =\frac{1-\tan \alpha \tan \beta }{1+\tan \alpha \tan \beta }
\end{align}$
Hence, the left side of the given expression is equal to the right side, which is
$\frac{\cos \left( \alpha +\beta \right)}{\cos \left( \alpha -\beta \right)}=\frac{1-\tan \alpha \tan \beta }{1+\tan \alpha \tan \beta }$.
