Finding $\frac{\cot\gamma}{\cot \alpha+\cot\beta}$, given $a^2+b^2=2019c^2$ This is a question that appeared in the $2018$ Southeast Asian Mathematical Olympiad:

In a triangle with sides $a,b,c$ opposite angles $\alpha,\beta,\gamma$, it is known that $$a^2+b^2=2019c^2$$ Find $$\frac{\cot\gamma}{\cot\alpha+\cot\beta} $$

Well, by the Sine Law we have $$\sin^2\alpha+\sin^2\beta=2019\sin^2\gamma$$ and by the Cosine Law, $$\cos\gamma=\frac{a^2+b^2-c^2}{2ab} = \frac{1009c^2}{ab}$$ I’m stuck here. I tried to convert everything in our target expression to sines and cosines, but that makes the expression more complicated. I guess we can use the fact that $\cot\gamma=-\cot(\alpha+\beta)$.
How can you tackle this question? (also, apparently there are no worked solutions online)
 A: Here is the complete solution from AoPS.
Continuing from Aqua's answer, we know that $\sin A = \frac{a}{2R}, \sin B = \frac{b}{2R}, \sin C = \frac{c}{2R}$. Furthermore, as you have found, $\cos C = \frac{1009c^2}{ab}$. Substituting these values in gives:
$$\frac{\cot\gamma}{\cot \alpha+\cot\beta} = \frac{\cos\gamma \sin \alpha \sin \beta}{\sin ^2\gamma} = \frac{\frac{1009c^2}{ab} \cdot \frac{a}{2R} \cdot \frac{b}{2R}}{\frac{c^2}{4R}} = \frac{\frac{1009c^2}{ab} \cdot ab}{c^2} = \frac{1009c^2}{c^2} = \boxed{1009}.$$
A: \begin{align}
\frac{\cot\gamma}{\cot \alpha+\cot\beta} &= \frac{\cot\gamma \sin \alpha \sin \beta}{\cos \alpha \sin \beta+\sin \alpha \cos\beta}\\
\\
&= \frac{\cot\gamma \sin \alpha \sin \beta}{\sin (\alpha +\beta)}\\
\\
&= \frac{\cos\gamma \sin \alpha \sin \beta}{\sin ^2\gamma}\\
\end{align}
Now it should not be difficult to finish.
A: $$\frac{\cot\gamma}{\cot\alpha+\cot\beta}=\frac{\cos\gamma\sin\alpha\sin\beta}{\sin\gamma\sin(\alpha+\beta)}=\frac{\cos\gamma\sin\alpha\sin\beta}{\sin^2\gamma}=$$
$$=\frac{\frac{a^2+b^2-c^2}{2ab}\cdot\frac{a}{2R}\cdot\frac{b}{2R}}{\frac{c^2}{4R^2}}=\frac{\frac{2018c^2}{2ab}\cdot\frac{ab}{4R^2}}{\frac{ c^2}{4R^2}}=1009.$$
