5. Given that $\angle A, \angle B$ are acute angles, satisfying
$$
\frac{\sin A}{\sin B}=\cos (A+B) \text{. }
$$

Then the maximum value of $\tan A$ is
5. $\frac{\sqrt{2}}{4}$.
$$
\begin{array}{l}
\text { Given } \sin [(A+B)-B]=\cos (A+B) \cdot \sin B \\
\Rightarrow \sin (A+B) \cdot \cos B=2 \cos (A+B) \cdot \sin B \\
\Rightarrow \tan (A+B)=2 \tan B .
\end{array}
$$

Thus $\tan A=\tan [(A+B)-B]$
$$
\begin{array}{l}
=\frac{\tan (A+B)-\tan B}{1+\tan (A+B) \cdot \tan B} \\
=\frac{\tan B}{1+2 \tan ^{2} B}=\frac{1}{\cot B+2 \tan B} \\
\leqslant \frac{1}{2 \sqrt{2}}=\frac{\sqrt{2}}{4} .
\end{array}
$$

The equality holds if and only if $2 \tan B=\cot B$, i.e., $\tan B=\frac{\sqrt{2}}{2}$.