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