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Thread: How should I have solved this?

  1. #1
    Sep 2009
    North West England

    How should I have solved this?

    So just got my exam and there was this question on it that I just didn't know where to start, wondered if any of you could explain to me how it should have been done.

    Had to show that $\displaystyle (cosecx)/(cosecx+1)+(cosecx)/(cosecx-1)=50$ could be written as $\displaystyle sec^2x=25$

    I'm pretty sure that was the question, if its not possible then the + is probably a -after the first fraction.

    Also the next part was to solve $\displaystyle sec^2x = 25$ and all my friends got 2 answers and I got 4, does that sound right?

    Only identies we can use in this exam are

    $\displaystyle tan^2x+1=sec^2x$
    $\displaystyle cot^2x+1=cosec^2x$
    $\displaystyle sin^2x+cos^2x=1$
    and all the 1 overs for tan sin and cos.

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  2. #2
    e^(i*pi)'s Avatar
    Feb 2009
    West Midlands, England
    Throughout I will use $\displaystyle \csc(x) = cosec(x)$ because I'm lazy

    A good start will be to get a common denominator which will be $\displaystyle (\csc(x)+1)(\csc(x)-1)$.

    $\displaystyle \dfrac{\csc(x)(\csc(x)-1)}{(\csc(x)+1)} + \dfrac{\csc(x)(\csc(x)+1)}{\csc(x)-1} = \dfrac{\csc(x)(\csc(x)-1) + \csc(x)(\csc(x)+1)}{(\csc(x)-1)(\csc(x)+1)}$

    When you expand you get (note that the denominator is difference of two squares)

    $\displaystyle \dfrac{\csc^2(x)-1 + \csc^2(x)+1}{\csc^2(x)-1} = \dfrac{2\csc^2(x)}{\csc^2(x)-1}$

    Using the identities given we get $\displaystyle \dfrac{2\csc^2(x)}{\cot^2(x)} = \dfrac{2}{\sin^2(x)} \cdot \dfrac{\sin^2(x)}{\cos^2(x)}$

    since we're dividing by a fraction I flipped it and multiplied. Cancelling gives

    $\displaystyle \dfrac{2}{\cos^2(x)} = 2\sec^2(x)$

    Set this equal to 50 and divide by 2 to finish the question

    edit: don't want to give the whole working away so put some in a spoiler. Look if you want, tis no water off my back
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