# Another tan sum problem!

• Jun 3rd 2006, 12:59 AM
Joyce
help..

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

:o :o
• Jun 3rd 2006, 01:08 AM
malaygoel
Quote:

Originally Posted by Joyce
help..

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

:o :o

tan(A-B)= -tan(B-A) =-tan((B-C)+(C-A))
expand it and you will get the answer
:)
• Jun 3rd 2006, 01:25 AM
CaptainBlack
Quote:

Originally Posted by Joyce
help..

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

:o :o

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

Also don't post the same question twice.

RonL
• Jun 3rd 2006, 03:17 AM
Soroban
Hello, Joyce!

Quote:

$\displaystyle \tan(A-B) + \tan(B-C) + \tan(C-A)\;=\;\tan(A-B)$$\displaystyle \tan(B-C)\tan(C-A) Let \displaystyle P = A-B,\;\;Q = B-C,\;\;R = C-A We see that: \displaystyle P + Q + R \:= \:(A - B) + (B - C) + (C - A)\:=\:0 So we have: \displaystyle P + Q + R\:=\:0\quad\Rightarrow\quad P + Q\:=\:-R Take the tangent of both sides: \displaystyle \tan(P + Q)\;=\;\tan(-R) . . and we have: \displaystyle \frac{\tan(P) + \tan(Q)}{1 - \tan(P)\cdot\tan(Q)}\;=\;-\tan(R) Then: \displaystyle \tan(P) + \tan(Q)\;=\;-\tan(R)[1 - \tan(P)\tan(Q)] \;=\;$$\displaystyle -\tan(R) + \tan(P)\tan(Q)\tan(R)$

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

Therefore: $\displaystyle \tan(A-B) + \tan(B-A) + \tan(C-A) \;=\;\tan(A-B)$$\displaystyle \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 $\displaystyle A + B + C = 180^o$, then: $\displaystyle \tan A + \tan B + \tan C\;=\,\tan A\tan B\tan C$

And the proof is almost identical.