$ABC$ is a triangle with $a,b$ and $c$ are the opposite sides of the angles. Let $(a^2+b^2):c^2=2019:1$, find $(\cot A+\cot B+\cot C:\cot A +\cot B)$

Question

$ABC$ is a triangle with $a,b$ and $c$ are the opposite sides of the angles. Let $(a^2+b^2):c^2=2019:1$, find $(\cot A+\cot B+\cot C:\cot A +\cot B)$

My attempt: \begin{align*} a^2+b^2-2ab\cos C&=c^2\\ 2019-\dfrac{2ab\cos C}{c^2}&=1\\ 1009&=\dfrac{ab\cos C}{c^2} \end{align*}

Answer 1

$$\frac{\sum\limits_{cyc}\cot\alpha}{\cot\alpha+\cot\beta}=1+\frac{\frac{\cos\gamma}{\sin\gamma}}{\frac{\sin\gamma}{\sin\alpha\sin\beta}}=1+\frac{\cos\gamma\sin\alpha\sin\beta}{\sin^2\gamma}=$$ $$=1+\frac{\frac{a^2+b^2-c^2}{2ab}\cdot\frac{2S}{bc}\cdot\frac{2S}{ac}}{\frac{4S^2}{a^2b^2}}=\frac{1}{2}+\frac{a^2+b^2}{2c^2}=1010.$$

Answer 2

Hint:

Using cosine & sine formula of Triangle,

$$\dfrac{\cot C}{\cot B+\cot A}=\dfrac{\dfrac{(a^2+b^2-c^2)2R}{2abc}}{\dfrac{2R(c^2+a^2-b^2+b^2+c^2-a^2)}{2abc}}=\dfrac{a^2+b^2-c^2}{2c^2}=\dfrac12\cdot\dfrac{a^2+b^2}{c^2}-\dfrac12=?$$