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 January 7th, 2010, 03:06 PM #1 Newbie   Joined: Jan 2010 Posts: 1 Thanks: 0 trig Prove that if A, B, and C are the angles of any triangle then tanA+tanB+tanC= tanAtanBtanC
January 7th, 2010, 05:33 PM   #2
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Re: trig

Hello, rsir!

This is a classic proof . . .

$\displaystyle \text{We need this identity: }\;\tan(\alpha\,\pm\,\beta) \;=\; \frac{\tan\alpha\,\pm\,\tan\beta}{1\,\mp\, \tan\alpha\,\tan\beta}$

Quote:
 Prove that if $\displaystyle A,\, B,\,C$ are the angles of any triangle, then: $\displaystyle \tan A\,+\,\tan B\,+\,\tan C\:=\: \tan A\,\tan B\,\tan C$

$\displaystyle \text{We have: }\;A\,+\,B\,+\,C\;=\;180^\circ \qquad\Rightarrow\quad A\,+\,B \;=\;180^\circ\,-\,C$

$\displaystyle \qquad\quad\text{Then: }\qquad\quad \tan(A\,+\,B) \;=\;\tan(180^\circ\,-\,C)$

$\displaystyle \frac{\tan A\,+\,\tan B}{1\,-\,\tan A\,\tan B} \;=\;\frac{\overbrace{\tan180^\circ}^{\text{This is 0}} \,-\, \tan C}{1 \,+\, \underbrace{\tan180^\circ}_{\text{This is 0}}\tan C}$

$\displaystyle \frac{\tan A\,+\,\tan B}{1\,-\,\tan A\,\tan B} \;=\;\frac{-\tan C}{1}$

$\displaystyle \tan A\,+\,\tan B \;=\;-\tan C \,+\,\tan A\,\tan B\,\tan C$

$\displaystyle \tan A\,+\,\tan B \,+\,\tan C \;=\;\tan A\,\tan B\,\tan C$

ta-DAA!

Last edited by skipjack; May 7th, 2017 at 09:27 AM.

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