Trigonometric Identities in Fractions

Today I have another trigonometric identity. I also found a nice questions in modular arithmetic, but it will have to wait until tomorrow, since this trigonometry question is much better. The question is short, which means it has a long solution. Here it is, try before you peek:

A + B + C = pi (A, B, and C are in radians).
Prove:
(cot A + cot B) / (tan A + tan B) + (cot B + cot C) / (tan B + tan C) + (cot A + cot C) / (tan A + tan C) = 1

Yeah, that's all you need to prove. That given the first condition, this long sum equals 1. And no, you can't multiply everything by the denominators. That's not how you prove an identity (and besides that, I'm not sure you want to multiply three binomial terms).

All we're left to do is hard work. Lots of fun. Let's see what happens when you take each fraction and try to simplify it a little (I'll write cot A as cotA for simplicity, same for B and C):

First, let's start with the numerator:
cotA + cotB = cosA/sinA + cosB/sinB
Make a common denominator and add:
(cosAsinB + cosBsinA) / (sinAsinB)
The numerator here looks like a trigonometric identity which says:
sin(A + B) = sinAcosB + cosAsinB
So the numerator of the first fraction is:
sin(A + B) / (sinAsinB)

Now the denominator:
tanA + tanB = sinA/cosA + sinB/cosB
Again, make common denominator and add:
(sinAcosB + sinBcosA) / (cosAcosB)
Looks familiar, doesn't it?
sin(A + B) / (cosAcosB)

Now you have the numerator and the denominator of the first fraction. Let's see what we can get out of them:
sin(A + B)/(sinAsinB) / sin(A + B)/(cosAcosB)

As you can see, sin(A + B) cancels (thank god) and we're left with (cosAcosB)/(sinAsinB). As you should know, this equals to cotAcotB. Well, now we have something we can work with.

I'm not going to show work on the other two fractions, I just assume you'll understand how I make this transition now:
cotAcotB + cotBcotC + cotAcotC = 1

Now that's more like it. Let's take cotC out as a common factor. Considering that C = pi - A - B, this can make our life easier. Follow closely now:
cotAcotB + cotC(cotA + cotB)
cotAcotB + cot(pi - A - B)(cotA + cotB)
Since tan(pi - x) = -tan(x), cot(pi - x) = -cot(x), so:
cotAcotB + cot(A + B)(cotA + cotB)

tan(A + B) = (tanA + tanB) / (1 - tanAtanB), so cot(A + B) = (1 - tanAtanB)/(tanA + tanB).

cotAcotB - (1 - tanAtanB)/(tanA + tanB) * (1/tanA + 1/tanB)
Make a common denominator and add, again:
cotAcotB - (tanA + tanB)/(1 - tanAtanB) * (tanA + tanB)/(tanAtanB)

Luckily, we can cancel (tanA + tanB):
cotAcotB - (1 - tanAtanB)/(tanAtanB)
cotAcotB - 1/(tanAtanB) + (tanAtanB)/(tanAtanB)

The first fraction equals cotAcotB and the second one equals 1:
cotAcotB - cotAcotB + 1
= 1

Proved.

That was long, but worth it, wasn't it?

Tomorrow, if there's nothing better, we'll have some modular arithmetic.

Yours,
Nadav

nadavs