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Math Help - one more proof

  1. #1
    Senior Member
    Joined
    Aug 2009
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    349

    one more proof

    in this problem i have to prove

    the inverse of cot (2) + inverse of cot (3) = pi/4

    so i rewrote the problem in terms of inverse tan and got

    inverse of tan (1/2) + inverse of tan (1/3)

    after i get to this step im not sure where to go...i used triangles trying to figure out the angle measures and came up with some weird numbers...one triangle showed sin = 1/sqrt of 5 and cos = 2/sqrt 5. the other triangle shows sin x = 1/sqrt of 10 and cos x = 3/sqrt of 10? any help would be appreciated. thanks in advance
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  2. #2
    MHF Contributor
    Joined
    Nov 2008
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    France
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    1,458
    Hi

    \tan \left(\tan^{-1}\left(\frac12\right) + \tan^{-1}\left(\frac13\right)\right) = \frac{\tan \left(\tan^{-1}\left(\frac12\right)\right) + \tan \left(\tan^{-1}\left(\frac13\right)\right)}{1 - \tan \left(\tan^{-1}\left(\frac12\right)\right) \times \tan \left(\tan^{-1}\left(\frac13\right)\right)}

    \tan \left(\tan^{-1}\left(\frac12\right) +  \tan^{-1}\left(\frac13\right)\right) = \frac{\frac12 + \frac13}{1 - \frac12 \times \frac13}

    \tan \left(\tan^{-1}\left(\frac12\right) +  \tan^{-1}\left(\frac13\right)\right) = 1 = \tan\left(\frac{\pi}{4}\right)

    0 < \tan^{-1}\left(\frac12\right) < \frac{\pi}{2}

    0 < \tan^{-1}\left(\frac13\right) < \frac{\pi}{2}

    0 < \tan^{-1}\left(\frac12\right) + \tan^{-1}\left(\frac13\right) < \pi

    Therefore \tan^{-1}\left(\frac12\right) + \tan^{-1}\left(\frac13\right) = \frac{\pi}{4}
    Last edited by running-gag; April 25th 2010 at 09:15 AM. Reason: Precision added
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