This is a question that appeared in the $2018$ Southeast Asian Mathematical Olympiad:

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