<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] It shouldn't be that hard cause you have this two.
$\frac{\cos (\frac{A}{2})}{\sin (\frac{A}{2})}+\frac{\cos (\frac{B}{2})}{\sin (\frac{B}{2})}+\frac{\cos (\frac{C}{2})}{\sin (\frac{C}{2})}=\frac{\cos (\frac{A}{2})}{\sin (\frac{A}{2})}\times \frac{\cos (\frac{B}{2})}{\sin (\frac{B}{2})}\times \frac{\cos (\frac{C}{2})}{\sin (\frac{C}{2})}$
$A+B+C=180$
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