18. In the diagram below, the lengths of the three sides of the triangle are $a \mathrm{~cm}, b \mathrm{~cm}$ and $c \mathrm{~cm}$. It is given that $\frac{a^{2}+b^{2}}{c^{2}}=2011$. Find the value of $\frac{\cot C}{\cot A+\cot B}$.
18. Answer: 1005
By the laws of sine and cosine, we have
$$
\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c} \text { and } \quad \cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}
$$

Then
$$
\begin{aligned}
\frac{\cot C}{\cot A+\cot B} & =\frac{\cos C}{\sin C} \cdot \frac{1}{\frac{m \sin B+\cos B \sin A}{\sin A \sin B}} \\
& =\frac{\sin A \sin B \cos C}{\sin (A+B) \sin C} \\
& =\left(\frac{\sin A \sin B}{\sin ^{2} C}\right) \cos C \\
& =\left(\frac{\left(a b / c^{2}\right) \sin ^{2} C}{\sin ^{2} C}\right)\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right) \\
& =\frac{a^{2}+b^{2}-c^{2}}{2 c^{2}} \\
& =\frac{2011-1}{2}=1005 .
\end{aligned}
$$