9. In $\triangle A B C$, let $B C=a, C A=b, A B=c$. If $9 a^{2}+9 b^{2}-19 c^{2}=0$, then $\frac{\cot C}{\cot A+\cot B}=$ $\qquad$ .
$$
\begin{array}{l}
\frac{\cot C}{\cot A+\cot B}=\frac{\frac{\cos C}{\sin C}}{\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}}=\frac{\frac{\cos C}{\sin C}}{\frac{\sin C}{\sin A \sin B}}=\frac{\sin A \sin B \cos C}{\sin ^{2} C} \\
=\frac{a b}{c^{2}} \cdot \frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{a^{2}+b^{2}-c^{2}}{2 c^{2}}=\frac{5}{9} .
\end{array}
$$