In $\triangle A B C$, $A B$ is the longest side, $\sin A \sin B=\frac{2-\sqrt{3}}{4}$, then the maximum value of $\cos A \cos B$ is

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9. $\frac{2+\sqrt{3}}{4}$.

Given that $AB$ is the longest side, $\angle A$ and $\angle B$ are both acute angles.
$$
\cos A \cos B + \sin A \sin B = \cos (A - B) \leqslant 1, \cos A \cos B \leqslant 1 - \sin A \sin B = \frac{2+\sqrt{3}}{4} \text{, equality holds when }
$$
$\angle A = \angle B$, i.e., $\sin A = \sin B = \frac{\sqrt{6} - \sqrt{2}}{4}$.