2. Let $a$, $b$, and $c$ be the sides of $\triangle ABC$, and $a^{2}+b^{2}=m c^{2}$. If $\frac{\cot C}{\cot A+\cot B}=999$, then $m$ $=$ $\qquad$
2.1999 .
$$
\begin{array}{l}
\because \frac{\cot C}{\cot A+\cot B}=\frac{\frac{\cos C}{\sin C}}{\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}} \\
\quad=\frac{\sin A \cdot \sin B \cdot \cos C}{\sin C(\cos A \cdot \sin B+\sin A \cdot \cos B)} \\
\quad=\cos C \cdot \frac{\sin A \cdot \sin B}{\sin ^{2} C}=\frac{a^{2}+b^{2}-c^{2}}{2 a b} \cdot \frac{a b}{c^{2}} \\
\quad=\frac{m-1}{2}=999, \\
\therefore m=2 \times 999+1=1999 .
\end{array}
$$