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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. Tags trig Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post cr1pt0 Trigonometry 2 September 5th, 2013 06:11 PM mathgeek11 Trigonometry 5 April 9th, 2013 08:53 PM Chee Trigonometry 7 December 11th, 2011 09:50 AM IneedofHelp Trigonometry 1 October 17th, 2011 02:38 AM Bihzad Algebra 1 March 11th, 2009 01:48 PM

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