China highshool math contest question

Let A, B, and C be the angles of the triangle, let a, b, and c be the sides of the triangle, and let R be the circumradius

First, know the sine rule:

\(\frac{a}{\sin{A}} =\frac{b}{\sin{B}}=\frac{c}{\sin{C}} = 2R\)

Rearrange like so:

\(\sin A = \frac{a}{2R}\\\sin B = \frac{b}{2R} \\\sin C = \frac{c}{2R}\)

Also, know the cosine rule (slightly rearranged so I don't have to write as much):

\(\cos{A} = \frac{b^2+c^2-a^2}{2bc}\\\cos{B} = \frac{c^2+a^2-b^2}{2ac}\\cos{C} = \frac{a^2+b^2-c^2}{2ab}\)

Notice that\(\frac{\cot{c}}{\cot{a}+\cot{b}} = \frac{\frac{\cos{c}}{\sin{c}}}{\frac{\cos{a}}{\sin{a}}+\frac{\cos{b}}{\sin{b}}}\)

Now, replace:

\(\frac{\frac{a^2+b^2-c^2}{2ab\cdot\frac{c}{2R}}}{\frac{b^2+c^2-a^2}{2bc\cdot\frac{a}{2R}}+ \frac{a^2+c^2-b^2}{2ac\cdot\frac{b}{2R}}}\)

Multiply the numerator and denominator by \(\frac{2R}{2abc}\):

\(\frac{a^2+b^2-c^2}{b^2+c^2-a^2+a^2+c^2-b^2} = \frac{a^2+b^2-c^2}{2c^2}\)

Rearrange the given equation:

\(a^2+b^2=\frac{19}{9}c^2\)

And then replace:

\(\frac{\frac{19}{9}c^2-c^2}{2c^2} = \frac{\frac{10}{9}c^2}{2c^2} = \boxed{\frac{5}{9}}\)

(this literally took 40 minutes to write so I hope this helps)