<s>[INST] In a triangle ABC, prove that cot(A/2)+cot(B/2)+cot(C/2) =cot(A/2)cot(B/2)cot(C/2) In a triangle ABC, prove that $\cot \left ( \frac{A}{2} \right )+\cot \left ( \frac{B}{2} \right )+\cot \left ( \frac{C}{2} \right )=\cot \left ( \frac{A}{2} \right )\times \cot \left ( \frac{B}{2} \right )\times \cot \left ( \frac{C}{2} \right )$. I tried all identities I know but I have no idea how to proceed. 
 [/INST] Geometric Proof
Let $a,b,c$ be the length of the sides $BC,CA,AB$ respectively, $p$ be the semi-perimeter, $r$ be the radius of the incircle, and $S$ be the area. 
Notice that $$\cot\frac{A}{2}=\frac{p-a}{r},~~~\cot\frac{B}{2}=\frac{p-b}{r},~~~\cot\frac{C}{2}=\frac{p-c}{r}.$$
Hence, $$\cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}=\frac{p}{r}.$$
and $$\cot\frac{A}{2}\cot\frac{B}{2}\cot\frac{C}{2}=\frac{(p-a)(p-b)(p-c)}{r^3}.$$
Recall Heron's formula $$S=\sqrt{p(p-a)(p-b)(p-c)},$$ and another area formula$$S=pr.$$
 We obtain $$\frac{(p-a)(p-b)(p-c)}{r^3}=\frac{S^2}{pr^3}=\frac{(pr)^2}{pr^3}=\frac{p}{r}.$$
As a result, $$\cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}=\cot\frac{A}{2}\cot\frac{B}{2}\cot\frac{C}{2}.$$
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