Proving $\cot(A)\cot(B)+\cot(B)\cot(C)+\cot(C)\cot(A)=1$ I was stumped by another past-year question:

In $\triangle ABC$, prove that $$\cot(A)\cot(B)+\cot(B)\cot(C)+\cot(C)\cot(A)=1.$$

Here's what I have done so far: I tried to replace $C$, using $C=180^\circ-(A+B)$. But after doing this, I don't know how to continue.
I would be really grateful for some help on this, thanks!

$$\cot(A+B+C)=\frac{\cot(A)\cot(B)\cot(C)-(\cot(A)+\cot(B)+\cot(C))}{\cot(A)\cot(B)+\cot(C)\cot(B)+\cot(C)\cot(A)-1}$$
now use the fact that $\cot(\pi)$ is infinity and for that the denominator on the right hand side has to be 0
