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, \[\frac{\cos \alpha \cos \beta}{\cos \gamma} = \frac{\sqrt{3}}{2 - \sqrt{3}} = \frac{\sqrt{3} (2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})} = \boxed{2 \sqrt{3} + 3}.\]