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!
 A: $$\cot(A+B+C)=\frac{\cot(A)\cot(B)\cot(C)-(\cot(A)+\cot(B)+\cot(C))}{\cot(A)\cot(B)+\cot(C)\cot(B)+\cot(C)\cot(A)-1}$$
now use the fact that $\cot(\pi)$ is infinity and for that the denominator on the right hand side has to be 0
A: So, we want to show that
$$\cot(A)\cot(B)+(\cot(A)+\cot(B))\cot(\pi-A-B)=1$$
Remembering that $\cot$ is odd ($\cot(-u)=-\cot u$) and has period $\pi$ ($\cot(u+\pi)=\cot u$), we have
$$\cot(A)\cot(B)-(\cot(A)+\cot(B))\cot(A+B)=1$$
At this point, you might want to turn everything into sines and cosines, use the addition formulae, and combine what can be combined.
Alternatively, you can derive the addition formula
$$\cot(\theta+\varphi)=\frac{\cot\,\theta\cot\,\varphi-1}{\cot\,\theta+\cot\,\varphi}$$
from the addition formulae for $\sin$ and $\cos$ and then substitute into your original identity.
A: Its surprising that everyone else missed this out, but there's actually a very simple and elegant solution to this proof.
So although this question is more than a year old, I still decided to post my proof.

I begin with a simple equation:
$$\cot(2x+x)=\cot{3x}$$

Now applying the formula: $\cot(A+B)=\frac{\cot{A}\cot{B}-1}{\cot{A}+\cot{B}}$, we get:

$$\frac{\cot{2x}\cot{x}-1}{\cot{2x}+\cot{x}}=\cot{3x}$$
$$\cot{2x}\cot{x}-1=\cot{2x}\cot{3x}+\cot{3x}\cot{x}$$
$$\cot{x}\cot{2x}-\cot{2x}\cot{3x}-\cot{3x}\cot{x}=1$$

and you're done.
