1. Given $a, b, c$ are the side lengths of $\triangle ABC$, satisfying $a^{2}+b^{2}=2018 c^{2}$. Then the value of $(\cot A+\cot B) \tan C$ is $\qquad$
$-1 \cdot \frac{2}{2017}$.
Notice,
$$
\begin{array}{l}
(\cot A+\cot B) \tan C \\
=\frac{\sin (A+B)}{\sin A \cdot \sin B} \cdot \frac{\sin C}{\cos C} \\
=\frac{\sin ^{2} C}{\sin A \cdot \sin B \cdot \cos C} \\
=\frac{2 c^{2}}{a^{2}+b^{2}-c^{2}}=\frac{2}{2017} .
\end{array}
$$