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!

  • $\begingroup$ It would be better if u showed what have you tried or what are your ideas as to begin with a solution for this. $\endgroup$
    – Bhargav
    Sep 7, 2011 at 8:10
  • $\begingroup$ @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? $\endgroup$
    – user38268
    Sep 7, 2011 at 8:32
  • $\begingroup$ @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. $\endgroup$
    – Sophia
    Sep 7, 2011 at 9:04
  • $\begingroup$ 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). $\endgroup$ Sep 7, 2011 at 9:56
  • $\begingroup$ 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. $\endgroup$
    – Srivatsan
    Sep 7, 2011 at 12:55

3 Answers 3



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


Remembering that $\cot$ is odd ($\cot(-u)=-\cot u$) and has period $\pi$ ($\cot(u+\pi)=\cot u$), we have


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


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:


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.


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