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Math Help - trigo problem

  1. #1
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    trigo problem

    Given  \alpha + \beta = 60 degree
    and m = \frac {\sqrt 3 + tan \alpha}{\sqrt 3 + tan\beta}

    Prove that tan\beta = \frac {2-m}{\sqrt 3 m}<br />
    Thanks in advance.
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  2. #2
    Super Member bigwave's Avatar
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    from

    <br />
m = \frac {\sqrt 3 + tan \alpha}{\sqrt 3 + tan\beta}<br />

    you can obtain

    \tan{\beta} = \frac{\sqrt{3} + \tan{\alpha}-\sqrt{3}m}{m}

    now you have

     \frac{\sqrt{3} + \tan{\alpha}-\sqrt{3}m}{m} = \frac{2-m}{\sqrt{3}{m}}
    Last edited by bigwave; December 24th 2009 at 01:09 AM.
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  3. #3
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    Quote Originally Posted by bigwave View Post
    \tan\left(\frac{\pi}{3}\right) = \sqrt{3}

    might help
     m = \frac {\sqrt 3 + tan (60-\beta)}{\sqrt 3 + tan \beta}
    =\frac{\sqrt 3 + \frac {\sqrt 3 - tan \beta}{1+\sqrt3 tan \beta}}{\sqrt 3 + tan \beta}
    = \frac {2 \sqrt 3 + 2 tan \beta}{\sqrt3 tan\beta (1+ \sqrt3 tan \beta)}
    ???
    don't know what to do next
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  4. #4
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    Grandad's Avatar
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    Hello cakeboby
    Quote Originally Posted by cakeboby View Post
    Given  \alpha + \beta = 60 degree
    and m = \frac {\sqrt 3 + tan \alpha}{\sqrt 3 + tan\beta}

    Prove that tan\beta = \frac {2-m}{\sqrt 3 m}<br />
    Thanks in advance.
    \alpha=60^o-\beta

    \Rightarrow \tan\alpha = \frac{\sqrt3-\tan\beta}{1+\sqrt3\tan\beta} (1)

    \frac{2-m}{m}=\frac{2}{m}-1
    =\frac{2(\sqrt3+\tan\beta)}{\sqrt3+\tan\alpha}-1

    =\frac{2(\sqrt3+\tan\beta)}{\sqrt3+\left(\dfrac{\s  qrt3-\tan\beta}{1+\sqrt3\tan\beta}\right)}-1 from (1)
    Simplify this. You'll find it comes to \sqrt3\tan\beta.

    Grandad
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  5. #5
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    thanks
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