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