6. In $\triangle A B C$, $\angle A \leqslant \angle B \leqslant \angle C$, if $\frac{\sin A+\sin B+\sin C}{\cos A+\cos B+\cos C}=\sqrt{3}$,
then the value of $\sin B+\sin 2 B$ is $\qquad$
6. $\sqrt{3}$.

Let $\alpha=\angle A-\frac{\pi}{3}, \beta=\angle B-\frac{\pi}{3}, \gamma=\angle C-\frac{\pi}{3}$.
Obviously, $\alpha+\beta+\gamma=0$.
From the given information, we have
$$
\begin{array}{l}
(\sin A-\sqrt{3} \cos A)+(\sin B-\sqrt{3} \cos B)+ \\
(\sin C-\sqrt{3} \cos C)=0 \\
\Rightarrow 2 \sin \left(A-\frac{\pi}{3}\right)+2 \sin \left(B-\frac{\pi}{3}\right)+ \\
2 \sin \left(C-\frac{\pi}{3}\right)=0 \\
\Rightarrow \sin \alpha+\sin \beta+\sin \gamma=0 .
\end{array}
$$

From $\sin \alpha+\sin \beta+\sin \gamma$
$$
\begin{array}{l}
=2 \sin \frac{\alpha+\beta}{2} \cdot \cos \frac{\alpha-\beta}{2}+2 \sin \frac{\gamma}{2} \cdot \cos \frac{\gamma}{2} \\
=-2 \sin \frac{\gamma}{2} \cdot \cos \frac{\alpha-\beta}{2}+2 \sin \frac{\gamma}{2} \cdot \cos \frac{\alpha+\beta}{2} \\
=2 \sin \frac{\gamma}{2}\left(\cos \frac{\alpha+\beta}{2}-\cos \frac{\alpha-\beta}{2}\right) \\
=-4 \sin \frac{\gamma}{2} \cdot \sin \frac{\alpha}{2} \cdot \sin \frac{\beta}{2} \\
\Rightarrow \sin \frac{\alpha}{2} \cdot \sin \frac{\beta}{2} \cdot \sin \frac{\gamma}{2}=0
\end{array}
$$
$\Rightarrow \sin \frac{\alpha}{2}, \sin \frac{\beta}{2}, \sin \frac{\gamma}{2}$ must have one that is 0.
Noting that $\frac{\alpha}{2}, \frac{\beta}{2}, \frac{\gamma}{2} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, we know that one of $\frac{\alpha}{2}, \frac{\beta}{2}, \frac{\gamma}{2}$ must be 0.
It is easy to see that, regardless of which one is 0, combined with
$$
\alpha+\beta+\gamma=0 \text { and } \alpha \leqslant \beta \leqslant \gamma,
$$

we have $\beta=0$, i.e., $\angle B=\frac{\pi}{3}$.
Thus, $\sin B+\sin 2 B=\sqrt{3}$.