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Math Help - (cos^2x-sin^2x)/cos^2x+sinxcosx=1-tanx

  1. #1
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    (cos^2x-sin^2x)/cos^2x+sinxcosx=1-tanx

    Can someone PLEASEEEEEE help me prove this trig identity.

    I did it down to where I have
    LS:
    sinx/cosx

    RS:
    -sinx

    I am not sure where I went wrong,
    can someone please help
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    LHS:
    For numerator use formula

    a^2 - b^2 = (a-b)(a+b)

    So

    cos^2(x) - sin^2 (x) = (cos(x) - sin(x)) (cos(x) + sin(x))


    Denominator

    cos^2(x) +cos(x)sin(x) = cos(x) (cos(x) +sin(x))


    Divide numerator by denominator

    The things in bold get cancelled

    -------------------------------------

    One more thing try to avoid writing like thissssssss

    Adarsh
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  3. #3
    MHF Contributor
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    Talking

    Until you show your steps, I'm afraid we can't be sure where you went wrong, either. Sorry!

    Please reply with a clear listing of your work and reasoning so far. Thank you!
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  4. #4
    Like a stone-audioslave ADARSH's Avatar
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    Sorry to disappoint you SHOULD start a new thread for a new problem

    Take Care of Yourself & Rules
    Adarsh
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  5. #5
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    Allahabad,Uttarpradesh
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    Smile

    Quote Originally Posted by ADARSH View Post
    LHS:
    For numerator use formula

    a^2 - b^2 = (a-b)(a+b)

    So

    cos^2(x) - sin^2 (x) = (cos(x) - sin(x)) (cos(x) + sin(x))


    Denominator

    cos^2(x) +cos(x)sin(x) = cos(x) (cos(x) +sin(x))


    Divide numerator by denominator

    The things in bold get cancelled

    -------------------------------------

    One more thing try to avoid writing like thissssssss

    Adarsh
    .....ok that's fine. Now just divide the Numerator by denominator...
    and the answer comes ..........(cos(x)-sin(x))/cos(x)=1- tan(x);
    that's it. what is the problem?
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