tan(A-B)= -tan(B-A) =-tan((B-C)+(C-A))Originally Posted byJoyce

expand it and you will get the answer

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

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

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

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

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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.

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