6. 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 $\frac{\cot C}{\cot A+\cot B}=$
$\begin{array}{l}\text { 6. } \frac{5}{9} \text { Hint: } \frac{\cot C}{\cot A+\cot B}=\frac{\sin A \sin B}{\sin ^{2} C} \cdot \cos C=\frac{a b}{c^{2}} \cdot \frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{a^{2}+b^{2}-c^{2}}{2 c^{2}} \\ =\frac{9 a^{2}+9 b^{2}-9 c^{2}}{2 \cdot 9 c^{2}}=\frac{19 c^{2}-9 c^{2}}{18 c^{2}}=\frac{5}{9} \\\end{array}$