2. Given $\alpha, \beta \in\left(0, \frac{\pi}{2}\right), \sin \beta=2 \cos (\alpha+\beta) \cdot \sin \alpha\left(\alpha+\beta \neq \frac{\pi}{2}\right)$, then the maximum value of $\tan \beta$ is
2. $\frac{\sqrt{3}}{3}$ Detailed Explanation: From the condition, we have:
$$
\sin [(\alpha+\beta)-\alpha]=\sin (\alpha+\beta) \cos \alpha-\cos (\alpha+\beta) \sin \alpha=2 \cos (\alpha+\beta) \sin \alpha,
$$

which means $\tan (\alpha+\beta)=3 \tan \alpha$. Also, because
$$
\tan \beta=\tan [(\alpha+\beta)-\alpha]=\frac{\tan (\alpha+\beta)-\tan \alpha}{1+\tan (\alpha+\beta) \tan \alpha}=\frac{2 \tan \alpha}{1+3 \tan ^{2} \alpha}=\frac{2}{\frac{1}{\tan \alpha}+3 \tan \alpha} \leqslant \frac{\sqrt{3}}{3} .
$$