i have found a very lenghty trig method.......too lazy to post here method put $\cot(\frac{A}{2}) = \frac{p(p-a)}{S}$ where $p=\frac{a+b+c}{2}$ and so on ........ then use the formula $\cot(2\theta) = \frac {\cot^{2}(\theta) - 1}{2\cot(\theta)}$ and put the values in the expression......and the result will be followed

We use the following identities: $S = \frac12 ab \sin (\angle C)$, and the Law of Cosines. From the Law of Cosines, we have: $$a^2 = b^2 + c^2 - 2bc \cos (\angle A)$$$$b^2 = a^2 + c^2 - 2ac \cos (\angle B)$$$$c^2 = a^2 + b^2 - 2ab \cos (\angle C)$$Adding these up we have $$2ab \cos (\angle C)+ 2bc \cos (\angle A) + 2ac \cos (\angle B)= a^2 + b^2 + c^2.$$Since $ab = 2\frac{S}{\sin(\angle C)}$ and the same for the others we have $$2ab \cos (\angle C)+ 2bc \cos (\angle A) + 2ac \cos (\angle B) = 4S \cot (\angle C) + 4S \cot (\angle A) + 4 S\cot (\angle B) = a^2 + b^2 + c^2.$$Dividing both sides by $4S$ yields the desired equation.