4. If acute angles $\alpha, \beta$ satisfy
$$
\sin \alpha=\cos (\alpha+\beta) \cdot \sin \beta,
$$

then the maximum value of $\tan \alpha$ is $\qquad$
$\begin{array}{l}\text { 4. } \frac{\sqrt{2}}{4} \text {. } \\ \text { Given } \sin \alpha=\cos (\alpha+\beta) \cdot \sin \beta \\ =\sin (\alpha+\beta-\beta) \\ =\sin (\alpha+\beta) \cdot \cos \beta-\cos (\alpha+\beta) \cdot \sin \beta \\ \Rightarrow \tan (\alpha+\beta)=2 \tan \beta \\ \Rightarrow \tan \alpha=\frac{\tan (\alpha+\beta)-\tan \beta}{1+\tan (\alpha+\beta) \cdot \tan \beta} \\ =\frac{\tan \beta}{1+2 \tan ^{2} \beta}=\frac{1}{2 \tan \beta+\frac{1}{\tan \beta}} \leqslant \frac{\sqrt{2}}{4} .\end{array}$