### Instruction: Let $a$, $b$, $c$ be the three sides of a triangle, and let $\alpha$, $\beta$, $\gamma$ be the angles opposite them. If $a^2+b^2=1989c^2$, find the value of
\[\frac{\cot \gamma}{\cot \alpha+\cot \beta}.\].### Response: This problem involves trigonometry and geometry, so I want to draw a picture of the triangle to help me visualize the situation.
I label the sides and angles as given, and I also mark the altitude from $C$ to $AB$ as $h$.
I know that $\cot \gamma=\frac{a^2+b^2-c^2}{4[ABC]}$, where $[ABC]$ is the area of the triangle.
Similarly, I know that $\cot \alpha=\frac{b^2+c^2-a^2}{4[ABC]}$ and $\cot \beta=\frac{a^2+c^2-b^2}{4[ABC]}$.
So, the expression I want to find is $\frac{\frac{a^2+b^2-c^2}{4[ABC]}}{\frac{b^2+c^2-a^2}{4[ABC]}+\frac{a^2+c^2-b^2}{4[ABC]}}$.
I can simplify this by canceling out the $4[ABC]$ in the numerator and denominator, and I get $\frac{a^2+b^2-c^2}{2c^2}$.
Now, I notice that the given condition $a^2+b^2=1989c^2$ can help me simplify this further.
I can substitute $a^2+b^2$ for $1989c^2$ in the numerator, and I get $\frac{1989c^2-c^2}{2c^2}=\frac{1988c^2}{2c^2}=994$.
So, the final answer is 994.