Let $\alpha,$ $\beta,$ and $\gamma$ be three angles such that $\alpha + \beta + \gamma = \pi.$  If we are given that $\tan \alpha \tan \beta = \csc \frac{\pi}{3},$ then determine $\frac{\cos \alpha \cos \beta}{\cos \gamma}.$
First, $\tan \alpha \tan \beta = \csc \frac{\pi}{3} = \frac{2}{\sqrt{3}}.$  Then
\[\sin \alpha \sin \beta = \frac{2}{\sqrt{3}} \cos \alpha \cos \beta.\]Now, from the angle addition formula,
\begin{align*}
\cos \gamma &= \cos (\pi - \alpha - \beta) \\
&= -\cos (\alpha + \beta) \\
&= \sin \alpha \sin \beta - \cos \alpha \cos \beta \\
&= \frac{2}{\sqrt{3}} \cos \alpha \cos \beta - \cos \alpha \cos \beta \\
&= \frac{2 - \sqrt{3}}{\sqrt{3}} \cos \alpha \cos \beta.
\end{align*}Therefore,
\[\frac{\cos \alpha \cos \beta}{\cos \gamma} = \frac{\sqrt{3}}{2 - \sqrt{3}} = \frac{\sqrt{3} (2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})} = \boxed{2 \sqrt{3} + 3}.\]