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!

• It would be better if u showed what have you tried or what are your ideas as to begin with a solution for this. Sep 7, 2011 at 8:10
• @Sophia Please post questions showing that you have put some effort into trying to solve the problem. By the way are you posting this question in preparation for your SPM?
– user38268
Sep 7, 2011 at 8:32
• @D B Lim: Nope, I am only doing them for my school's final examination. I will only have my SPM next year. As for showing effort, I have been thinking, asking friends, looking up info about the topic, but still don't know how to continue. Sep 7, 2011 at 9:04
• What they meant was that you show what algebra you've tried to solve this problem (i.e., what you've tried writing on paper). Sep 7, 2011 at 9:56
• Not an answer, but you might have seen the identity: $\tan A + \tan B + \tan C = \tan A \cdot \tan B \cdot \tan C$. These two are really the same. Sep 7, 2011 at 12:55

$$\cot(A+B+C)=\frac{\cot(A)\cot(B)\cot(C)-(\cot(A)+\cot(B)+\cot(C))}{\cot(A)\cot(B)+\cot(C)\cot(B)+\cot(C)\cot(A)-1}$$

now use the fact that $\cot(\pi)$ is infinity and for that the denominator on the right hand side has to be 0

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.

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.