<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] Here’s a somewhat weird proof using inverse trigonometric functions  
If  $$A+B+C=π$$ then  $$\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=\frac{π}{2}$$
Let  $$\frac{A}{2}=tan^{-1} x, \>\frac{B}{2}=tan^{-1} y ,\> \frac{C}{2}=tan^{-1} z$$ 
Now,
$$tan^{-1} x+ tan^{-1} y+ tan^{-1} z=  π/2$$
$$tan^{-1} x+ tan^{-1} y= cot^{-1} z$$
Because $$tan^{-1} z+ cot^{-1} z=   \frac{π}{2}$$
Now,
$$tan^{-1} \frac{x+y}{1-xy}= cot^{-1} z$$
$$cot^{-1} \frac{1-xy}{x+y}= cot^{-1} z$$
$$\frac{1-xy}{x+y}=z$$
$$1=xy+yz+zx$$
$$\frac{1}{xyz}=  \frac{1}{z}+ \frac{1}{x}+\frac{1}{y}$$
Now,we have
$$cot \frac{A}{2} cot \frac{B}{2} cot \frac{C}{2}= cot \frac{C}{2}+cot \frac{A}{2}+cot \frac{B}{2}$$
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