6. In $\triangle A B C$, $\sin ^{2} A+\sin ^{2} C=2018 \sin ^{2} B$, then $\frac{(\tan A+\tan C) \tan ^{2} B}{\tan A+\tan B+\tan C}=$
Solve $\frac{2}{2017}$.
Analysis Since $\sin ^{2} A+\sin ^{2} C=2018 \sin ^{2} B$, we have $a^{2}+c^{2}=2018 b^{2}$. Note that:
$$
\tan A+\tan B+\tan C=\tan A \cdot \tan B \cdot \tan C
$$

Therefore,
$$
\begin{aligned}
\frac{(\tan A+\tan C) \tan ^{2} B}{\tan A+\tan B+\tan C} & =\frac{(\tan A+\tan C) \tan ^{2} B}{\tan A \cdot \tan B \cdot \tan C}=\left(\frac{1}{\tan A}+\frac{1}{\tan C}\right) \tan B \\
& =\frac{\sin ^{2} B}{\sin A \cdot \sin C} \cdot \frac{1}{\cos B}=\frac{b^{2}}{a c}\left(\frac{2 a c}{a^{2}+c^{2}-b^{2}}\right) \\
& =\frac{2 b^{2}}{2018 b^{2}-b^{2}}=\frac{2}{2017}
\end{aligned}
$$