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

    After you expand the top bracket you will realise the signs change then u should be able to cancel

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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

    https://www.symbolab.com/solver/step...7B2%7Dx%7D%3D1
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    (Original post by Rtdsv)
    2 - ( 1 - sin^2x) / (1 + sin^2x)

    After you expand the top bracket you will realise the signs change then u should be able to cancel

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    Thanks for replying, how would you expand it?
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    2-(1-sin^2)/1+sin^2
    2-1+ sin^2/1+sin^2
    1+sin^2/1+sin^2
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    (Original post by Super helper)
    2-(1-sin^2)/1+sin^2
    2-1+ sin^2/1+sin^2
    1+sin^2/1+sin^2
    hi, thanks for replying. I don't understand why -sin^2 changes to +sin^2
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    You basically have -1 outside the bracket so when expanding you get -1 x 1 which is -1 and -1 x -sin^2x which is +sin^2x since - x - give positive. Then 2-1+sin^2x

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    (Original post by man111111)
    hi, thanks for replying. I don't understand why -sin^2 changes to +sin^2
    2-(1-sin^2)
    2-1--sin^2. 2 minuses Makes a plus

    U can write it as 2+(-1(1-sin^2)) if that helps you see it
    The minus doesnt only only act on the 1
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    (Original post by Rtdsv)
    You basically have -1 outside the bracket so when expanding you get -1 x 1 which is -1 and -1 x -sin^2x which is +sin^2x since - x - give positive. Then 2-1+sin^2x

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    thank you so much for helping me.
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    (Original post by Super helper)
    2-(1-sin^2)
    2-1--sin^2. 2 minuses Makes a plus

    U can write it as 2+(-1(1-sin^2)) if that helps you see it
    The minus doesnt only only act on the 1
    Hi thanks for helping me.
 
 
 
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