3. 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 $\frac{\cot C}{\cot A+\cot B}=$

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$\frac{5}{9}$
3. 【Analysis and Solution】Solution 1: $\frac{\cot C}{\cot A+\cot B}=\frac{\frac{\cos C}{\sin C}}{\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}}=\frac{\sin A \sin B}{\sin ^{2} C} \cos C$ $=\frac{a b}{c^{2}} \cdot \frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{a^{2}+b^{2}-c^{2}}{2 c^{2}}=\frac{9\left(a^{2}+b^{2}-c^{2}\right)}{18 c^{2}}=\frac{19 c^{2}-9 c^{2}}{18 c^{2}}=\frac{5}{9}$.
Solution 2 In fact, in $\triangle A B C$, we have $\cot C=\frac{\left(a^{2}+b^{2}-c^{2}\right)}{4 S}$, similarly, we have the expressions for $\cot A, \cot B$, substituting them into $\frac{\cot C}{\cot A+\cot B}$, and using the condition $9 a^{2}+9 b^{2}-19 c^{2}=0$, we immediately get $\frac{\cot C}{\cot A+\cot B}=\frac{\left(a^{2}+b^{2}-c^{2}\right)}{2 c^{2}}=\frac{5}{9}$.