### Instruction: 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 $\frac{\cos \alpha \cos \beta}{\cos \gamma}.$.### Response: This problem involves some trigonometric identities and equations, so I will start by recalling some of the basic ones that might be useful.
The first one is that $\tan x = \frac{\sin x}{\cos x},$ which relates the tangent function to the sine and cosine functions.
The second one is that $\csc x = \frac{1}{\sin x},$ which is the reciprocal of the sine function.
The third one is that $\cos(\pi - x) = - \cos x,$ which is a property of the cosine function for supplementary angles.
Now, I will look at the given equation, $\tan \alpha \tan \beta = \csc \frac{\pi}{3}.$
I will use the first identity to rewrite this equation in terms of sine and cosine functions:

\[\frac{\sin \alpha}{\cos \alpha} \cdot \frac{\sin \beta}{\cos \beta} = \frac{1}{\sin \frac{\pi}{3}}\]
Then, I will simplify the right-hand side using the fact that $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}:$

\[\frac{\sin \alpha}{\cos \alpha} \cdot \frac{\sin \beta}{\cos \beta} = \frac{2}{\sqrt{3}}\]
Next, I will cross-multiply and rearrange the equation to get:

\[\sin \alpha \sin \beta = \frac{2}{\sqrt{3}} \cos \alpha \cos \beta\]
Now, I want to find a way to use the fact that $\alpha + \beta + \gamma = \pi,$ which implies that $\gamma = \pi - (\alpha + \beta).$
I will use the third identity to write $\cos \gamma$ in terms of $\cos(\alpha + \beta):$

\[\cos \gamma = \cos(\pi - (\alpha + \beta)) = - \cos(\alpha + \beta)\]
Then, I will use the angle sum formula for the cosine function to expand $\cos(\alpha + \beta):$

\[\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta\]
Therefore, I can write $\cos \gamma$ as:

\[\cos \gamma = - (\cos \alpha \cos \beta - \sin \alpha \sin \beta)\]
Now, I have an expression for $\cos \gamma$ in terms of $\cos \alpha,$ $\cos \beta,$ $\sin \alpha,$ and $\sin \beta.$
I also have an equation that relates $\sin \alpha \sin \beta$ to $\cos \alpha \cos \beta.$
So, I will substitute the equation into the expression and simplify:

\[\begin{aligned}
\cos \gamma &= - (\cos \alpha \cos \beta - \sin \alpha \sin \beta)\\
&= - \cos \alpha \cos \beta + \sin \alpha \sin \beta\\
&= - \cos \alpha \cos \beta + \frac{2}{\sqrt{3}} \cos \alpha \cos \beta\\
&= (\frac{2}{\sqrt{3}} - 1) \cos \alpha \cos \beta
\end{aligned}\]