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    show that (2 - cos^2 x) / (1 + sin^2 x) is equal to 1

    I have substituted cos^2 x by (1 - sin^2 x) but I don't know what to do next
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    what past paper is this from?
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    (Original post by man111111)
    show that (2 - cos^2 x) / (1 + sin^2 x) is equal to 1

    I have substituted cos^2 x by (1 - sin^2 x) but I don't know what to do next
    Hi,

    I did for LHS:

    (2 - cos^2 x) / (1+ 1 - cos^2 x)

    So, (2 - cos^2 x) / ( 2 - cos ^2 x)

    which equals 1

    If you substitute cos^2 x by (1 - sin^2 x):

    [ 2 - (1 - sin^2 x)] / (1 + sin^2 x)

    So, ( 1 + sin^2 x ) / (1+ sin^2 x )
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    (Original post by XavierD)
    Hi,

    I did for LHS:

    (2 - cos^2 x) / (1+ 1 - cos^2 x)

    So, (2 - cos^2 x) / ( 2 - cos ^2 x)

    which equals 1

    If you substitute cos^2 x by (1 - sin^2 x):

    [ 2 - (1 - sin^2 x)] / (1 + sin^2 x)

    So, ( 1 + sin^2 x ) / (1+ sin^2 x )
    hi thanks for replying,
    I don't understand why [2 - (1 - sin^2 x)] gives you 1+sin^2x

    In addition, I don't understand why [1+ (1 - cos^2 x)] gives you 2 - cos^2 x
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    u need the rule

    sin2x + cos2x = 1

    so we got
    2-cos2x / 1+ sin2x

    we can substitute cos2x with 1-sin2x so we would get

    2 - (1-sin2x)/1+sin2x {I think u forgot that u have to put brackets in}

    so we would then get

    2 - 1 + sin2x / 1 + sin2x

    which equals to

    1 + sin2x / 1 + sin2x

    which then equals to
    1

    I think u forgot the fact that you had to put the 1 - sin2x inside a bracket.
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    (Original post by man111111)
    hi thanks for replying,
    I don't understand why [2 - (1 - sin^2 x)] gives you 1+sin^2x

    In addition, I don't understand why [1+ (1 - cos^2 x)] gives you 2 - cos^2 x
    I just opened the brackets,

    for: 2 - ( 1 - sin^2 x)

    equals: 2 - 1 + sin^2 x

    so, 1 +sin^2 x

    for: 1 + ( 1 - cos^2 x)

    opening the brackets: 1 + 1 - cos^2 x

    so, 2 - cos^2 x

    Please let me know if this helps
 
 
 
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