<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] Applying $\cot(x+y)=\dfrac{\cot x\cot y-1}{\cot x+\cot y}$ twice,
$$\cot(x+y+z)=\frac{\sum\cot x\cot y-1}{\prod\cot x-\sum\cot x}$$
Here $x=\dfrac A2$ etc. so that $x+y+z=\dfrac\pi2\implies\cot(x+y+z)=?$
</s>