7. In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively. If $a^{2}+b^{2}=2019 c^{2}$, then
$$
\frac{\cot C}{\cot A+\cot B}=
$$
$\qquad$ .
7. 1 009.

According to the problem, we have
$$
\begin{array}{l}
\frac{\cot C}{\cot A+\cot B}=\frac{\sin A \cdot \sin B \cdot \cos C}{\sin ^{2} C} \\
=\frac{a b}{c^{2}} \cdot \frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{2019 c^{2}-c^{2}}{2 c^{2}} \\
=1009 .
\end{array}
$$