Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
It would be better if u showed what have you tried or what are your ideas as to begin with a solution for this. –  Bhargav Sep 7 '11 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 '11 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. –  Sophia Sep 7 '11 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). –  J. M. Sep 7 '11 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. –  Srivatsan Sep 7 '11 at 12:55

2 Answers 2

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

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

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.