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Math Help - trig problems:

  1. #1
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    trig problems:

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

    2.if  \tan A=n \tan B ,then find the maximum value of  \tan^2 (A-B)
    Last edited by earthboy; October 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) tan(A-B)=\frac{tanA-tanB}{1+tanAtanB}.

    Now let we divide (1) by tanB:

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

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

    Putting (3) in (2):

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

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

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

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

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

    Or:

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

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

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

    Therefor max\left \{tan^2(A-B)  \right \}=\frac{(n-1)^2}{4n}.
    Last edited by Also sprach Zarathustra; November 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:

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

    to this:

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

    Concerning the denominator...

    Shouldn't it be:

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

    Or

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

    ?
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