2021/072) if in a triangle $a^2+b^2= \frac{19}{9}c^2$ evaluate $\frac{\cot\,C}{\cot\, A + \cot \, B}$

We are given
$a^2+b^2= \frac{19}{9}c^2$
Using the above and law of cosine we get
$2ab\cos\, C = a^2+b^2-c^2 = \frac{19}{9}c^2 - c^2 = \frac{10}{9}c^2$
or $ab\cos\, C = \frac{5c^2}{9}\cdots(1)$

Further
$\cot\, A +\cot\, B= \frac{\cos\, A}{\sin\, A} +  \frac{\cos\, B}{\sin\, B}$
$= \frac{\cos\, A\sin\, B +  \sin\,A\cos\, B}{\sin \, A\sin\, B}$
$= \frac{\sin(A+B)}{\sin \, A\sin\, B}$
$= \frac{\sin(\pi-C)}{\sin \, A\sin\, B}$ as $A+B+C=\pi$
$= \frac{\sin\,C}{\sin \, A\sin\, B}$

Hence $\frac{\cot\,C}{\cot\, A + \cot \, B} = \frac{\cos\, C \sin\, A\sin\, B}{\sin ^2C}$
$=\frac{ab\cos\,C}{c^2}$ (using law of sin)
$=\frac{5}{9}$