10. Let $a$, $b$, $c$ be the three sides of a triangle, and $\alpha$, $\beta$, $\gamma$ be the angles opposite these sides, respectively. If $a^{2}+b^{2}=1989 c^{2}$, find the value of $\frac{\cot \gamma}{\cot \alpha+\cot \beta}$.
$\begin{array}{l}\text { 10. } \frac{\cot \gamma}{\cot \alpha+\cot \beta}=\frac{\cos \gamma}{\sin \gamma} \cdot \frac{1}{\frac{\sin \beta \cos \alpha+\cos \beta \sin \alpha}{\sin \alpha \sin \beta}}=\frac{\cos \gamma \sin \alpha \sin \beta}{\sin \gamma \sin (\alpha+\beta)}=\frac{\cos \gamma \sin \alpha \sin \beta}{\sin ^{2} \gamma} \\ =\cos \gamma \cdot \frac{\sin \alpha}{\sin \gamma} \cdot \frac{\sin \beta}{\sin \gamma}=\frac{a^{2}+b^{2}-c^{2}}{2 a b} \cdot \frac{a}{c} \cdot \frac{b}{c}=\frac{1989 c^{2}-c^{2}}{2 c^{2}}=994 . \\\end{array}$