7 . In $\triangle \mathbf{A B C}$, the sides opposite to angles $\mathbf{A} 、 \mathbf{B} 、 \mathbf{C}$ are $a 、 \mathbf{b} 、 \mathbf{c}$ respectively. If $a^{2}+b^{2}=2019 c^{2}$, then $\frac{\cot C}{\cot A+\cot B}=$
【Answer】 1009
【Analysis】【Detailed Explanation】

From the problem, we get
$$
\frac{\cot C}{\cot A+\cot B}=\frac{\sin A \sin B \cos C}{\sin ^{2} C}=\frac{a b}{c^{2}} \cdot \frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{2019 c^{2}-c^{2}}{2 c^{2}}=1009
$$

Therefore, the answer is: 1009