Another tan sum problem!

**Joyce** · June 3rd 2006, 12:59 AM

help..

tan(A-B)+tan(B-C)+tan(C-A)=tan(A-B)tan(B-C)tan(C-A)

**malaygoel** · June 3rd 2006, 01:08 AM

Originally Posted by Joyce

help..

tan(A-B)+tan(B-C)+tan(C-A)=tan(A-B)tan(B-C)tan(C-A)

tan(A-B)= -tan(B-A) =-tan((B-C)+(C-A))
expand it and you will get the answer

**CaptainBlack** · June 3rd 2006, 01:25 AM

Originally Posted by Joyce

help..

tan(A-B)+tan(B-C)+tan(C-A)=tan(A-B)tan(B-C)tan(C-A)

This is a new question it belongs in a new thread.

Also don't post the same question twice.

RonL

**Soroban** · June 3rd 2006, 03:17 AM

Hello, Joyce!

$\tan(A-B) + \tan(B-C) + \tan(C-A)\;=\;\tan(A-B)$ $\tan(B-C)\tan(C-A)$

Let $P = A-B,\;\;Q = B-C,\;\;R = C-A$

We see that: $P + Q + R \:= \<img src=$ A - B) + (B - C) + (C - A)\:=\:0" alt="P + Q + R \:= \

A - B) + (B - C) + (C - A)\:=\:0" />

So we have: $P + Q + R\:=\:0\quad\Rightarrow\quad P + Q\:=\:-R$

Take the tangent of both sides: $\tan(P + Q)\;=\;\tan(-R)$

. . and we have: $\frac{\tan(P) + \tan(Q)}{1 - \tan(P)\cdot\tan(Q)}\;=\;-\tan(R)$

Then: $\tan(P) + \tan(Q)\;=\;-\tan(R)[1 - \tan(P)\tan(Q)] \;=\;$ $-\tan(R) + \tan(P)\tan(Q)\tan(R)$

. . Hence: $\tan(P) + \tan(Q) + \tan(R)\;=\;\ran(P)\tan(Q)\tan(R)$

Therefore: $\tan(A-B) + \tan(B-A) + \tan(C-A) \;=\;\tan(A-B)$ $\tan(B - C)\tan(C - A)$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

How did I come up with this proof?

I saw that this problem was similar to another (classic) problem:
. . If $A + B + C = 180^o$ , then: $\tan A + \tan B + \tan C\;=\,\tan A\tan B\tan C$

And the proof is almost identical.

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