Example 13 Given $\alpha, \beta \in\left(0, \frac{\pi}{2}\right)$, and $\sin \beta=2 \cos (\alpha+\beta) \cdot \sin \alpha\left(\alpha+\beta \neq \frac{\pi}{2}\right)$, find the maximum value of $\tan \beta$.

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Solve: From $2 \cos (\alpha+\beta) \cdot \sin \alpha=\sin \beta=\sin [(\alpha+\beta)-\alpha]=\sin (\alpha+\beta) \cdot \cos \alpha - \cos (\alpha+\beta) \cdot \sin \alpha$, we have $\sin (\alpha+\beta) \cdot \cos \alpha=3 \cos (\alpha+\beta) \cdot \sin \alpha$,
which means $\tan (\alpha+\beta)=3 \tan \alpha$.
$$
\begin{aligned}
\tan \beta & =\tan [(\alpha+\beta)-\alpha]=\frac{\tan (\alpha+\beta)-\tan \alpha}{1+\tan (\alpha+\beta) \cdot \tan \alpha} \\
& =\frac{2}{\frac{1}{\tan \alpha}+3 \tan \alpha} \leqslant \frac{2}{2 \sqrt{\frac{1}{\tan \alpha} \cdot 3 \tan \alpha}}=\frac{\sqrt{3}}{3},
\end{aligned}
$$

where the equality holds if and only if $\frac{1}{\tan \alpha}=3 \tan \alpha$, i.e., $\tan \alpha= \pm \frac{\sqrt{3}}{3}$. 
Therefore, when $\alpha=\frac{\pi}{6} \in\left(0, \frac{\pi}{2}\right)$, $\tan \alpha=\frac{\sqrt{3}}{3}$, the maximum value of $\tan \beta$ is $\frac{\sqrt{3}}{3}$, at this time $\beta=\frac{\pi}{6} \in\left(0, \frac{\pi}{2}\right)$.