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Results day: make sure you know what to expect before you get your grades. Here's how to be ready. 22-07-2016
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    http://store.aqa.org.uk/qual/gce/pdf...W-QP-JAN11.PDF
    http://store.aqa.org.uk/qual/gce/pdf...W-MS-JAN11.PDF

    7B) Unsure on the last stage of this question as I'm left with:

    -2cosec^2x/-cot^2x=50

    to get to Sec^2x=25

    thanks
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    cosec^2x = 1 + cot^2x

    so you have -2(1+cot^2x)/-cot^2x=50

    You can get rid of the negatives as they are on both the top and bottom

    2((sin^2x+cos^2x)/sin^2x)/cot^2x=50

    2(1/sin^2x)/cos^2x/sin^2x=50
    The sin^2x cancel out to give you:
    2 x1/cos^2x=50
    therefore... 2sec^2x=50
    sec^2x=25

    Hope that makes sense

    EDIT: You could also go from
    2cosec^2x/cot^2x to 2(1/sin^2x)/cos^2x/sin^2x

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