help..

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

Results 1 to 4 of 4

- June 3rd 2006, 12:59 AM #1

- Joined
- Jun 2006
- Posts
- 12

- June 3rd 2006, 01:08 AM #2

- June 3rd 2006, 01:25 AM #3

- Joined
- Nov 2005
- From
- someplace
- Posts
- 14,972
- Thanks
- 4

- June 3rd 2006, 03:17 AM #4

- Joined
- May 2006
- From
- Lexington, MA (USA)
- Posts
- 11,740
- Thanks
- 645

Hello, Joyce!

We see that: A - B) + (B - C) + (C - A)\:=\:0" alt="P + Q + R \:= \A - B) + (B - C) + (C - A)\:=\:0" />

So we have:

Take the tangent of both sides:

. . and we have:

Then:

. . Hence:

Therefore:

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

How did I come up with this proof?

I saw that this problem was similar to another (classic) problem:

. . If , then:

And the proof is almost identical.