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Thread: trig problems:

  1. #1
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    trig problems:

    1. prove that $\displaystyle \frac{ \cos A +1-\sin A }{ \cos A - 1+\sin A} = \frac{1+\cos A}{\sin A} $

    2.if $\displaystyle \tan A=n \tan B $ ,then find the maximum value of $\displaystyle \tan^2 (A-B) $
    Last edited by earthboy; Oct 31st 2010 at 10:53 PM.
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  2. #2
    MHF Contributor harish21's Avatar
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    what have you done? show your work.
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  3. #3
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    yup!got the first one...but have no idea on the second one.
    please help !
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    Let we start with the trig. formula:

    (1) $\displaystyle tan(A-B)=\frac{tanA-tanB}{1+tanAtanB}$.

    Now let we divide (1) by $\displaystyle tanB$:

    (2) $\displaystyle \frac{tan(A-B)}{tanB}=\frac{\frac{tanA}{tanB}-1}{1+tanAtanB}$

    We given that: (3) $\displaystyle \frac{tanA}{tanB}=n$

    Putting (3) in (2):

    $\displaystyle \frac{tan(A-B)}{tanB}=\frac{n-1}{\frac{1}{tanB}+ntanB}$

    Now I will assume that $\displaystyle tanB>0$, and will sign $\displaystyle tanB=t$.

    Let us now look at: $\displaystyle \frac{1}{tanB}+ntanB=\frac{1}{t}+nt$

    Using: Inequality of arithmetic and geometric means - Wikipedia, the free encyclopedia

    $\displaystyle \frac{\frac{1}{t}+nt}{2}\geq \sqrt{n}$

    Or:

    $\displaystyle {\frac{1}{t}+nt\geq 2\sqrt{n}$

    Hence $\displaystyle min\left \{ \frac{1}{tanB}+ntanB \right \}= 2\sqrt{n}$

    Maximum value of $\displaystyle tan(A-B)=\frac{n-1}{ 2\sqrt{n}}$

    Therefor $\displaystyle max\left \{tan^2(A-B) \right \}=\frac{(n-1)^2}{4n}$.
    Last edited by Also sprach Zarathustra; Nov 1st 2010 at 01:29 PM.
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  5. #5
    MHF Contributor Unknown008's Avatar
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    I would like to know how you went from this:

    $\displaystyle \dfrac{tan(A-B)}{tanB}=\dfrac{\frac{tanA}{tanB}-1}{1+tanAtanB}$

    to this:

    $\displaystyle \dfrac{tan(A-B)}{tanB}=\dfrac{n-1}{\frac{1}{tanB}+ntanB}$

    Concerning the denominator...

    Shouldn't it be:

    $\displaystyle \dfrac{tan(A-B)}{tanB}=\dfrac{n-1}{1+ntan^2B}$

    Or

    $\displaystyle \dfrac{tan(A-B)}{tanB}=\dfrac{\frac{n-1}{tanB}}{\frac{1}{tanB}+ntanB}$

    ?
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