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Math Help - trig help / simplifying

  1. #1
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    trig help / simplifying

    Given that  tan^{2}K = 2secK

    a) find the value of sec k.
    (b) deduce that cosk=√2−1


     tan^{2}k = 2secK

     sec^2k-1 = 2secK

     sec^2K-2secK-1 = 0

     secK = 1+ \sqrt{2}

    b)  \frac{1}{cosk} = 1+\sqrt{2}

     cosK = \frac{1}{1 + \sqrt{2}}

    I don't get the right answer, if I rationalise the denominator by multiplying top and bottom by  1 - \sqrt{2} I still don't get  cosK = \sqrt{2} -1
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  2. #2
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    Quote Originally Posted by Tweety View Post
    Given that  tan^{2}K = 2secK

    a) find the value of sec k.
    (b) deduce that cosk=√2−1


     tan^{2}k = 2secK

     sec^2k-1 = 2secK

     sec^2K-2secK-1 = 0

     secK = 1+ \sqrt{2}

    b)  \frac{1}{cosk} = 1+\sqrt{2}

     cosK = \frac{1}{1 + \sqrt{2}}

    I don't get the right answer, if I rationalise the denominator by multiplying top and bottom by  1 - \sqrt{2} I still don't get  cosK = \sqrt{2} -1
    \frac{1}{1+\sqrt{2}}\cdot\frac{1-\sqrt{2}}{1-\sqrt{2}}=\frac{1-\sqrt{2}}{1-(\sqrt{2})^2}=\frac{1-\sqrt{2}}{1-2}=\frac{1-\sqrt{2}}{-1}=\sqrt{2}-1
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  3. #3
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    what mistake i have done????????????please help
    <br />

    " alt="tan^{2}k = 2secK
    " />

    <br />
    " alt="sec^2k-1 = 2secK
    " />


    " alt="sec^2K-2secK-1 = 0
    " />
    after that if we take \sec k = x
    then
    x^2-2x-1=0
    x=1
    \sec x=1









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  4. #4
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    jashansinghal: your mistake is having the color tags inside of the math tags in LaTex...


    Oh, you mean the problem?
    x^2-2x-1=0
    x=1
    \sec x=1
    I don't know why you got x = 1 from the quadratic. You need to use the quadratic formula.
    x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-1}}{2(1)}
    x = \frac{2 \pm \sqrt{4 + 4}}{2}
    x = \frac{2 \pm 2\sqrt{2}}{2}
    x = 1 \pm \sqrt{2}

    Reject x = 1 - \sqrt{2} because secent of an angle cannot be in between -1 and 1.
    x = 1 + \sqrt{2}
    \sec k = 1 + \sqrt{2}


    01
    Last edited by yeongil; July 24th 2009 at 07:45 PM.
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  5. #5
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    okkk sorry
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