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Math Help - trig

  1. #1
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    trig

    we have 2 secx= cosx/1-sinx+cosx/1+sinx
    we must prove that it is equal
    2secx=cosx(1-sinx)+cosx(1+sinx)/1-sinx+1+sinx

    which give me
    cosx-cosxsinx+cosx+cosxsinx/sinx^2
    from this point i am lost have i make mistake ??
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by gregorio View Post
    we have 2 secx= cosx/1-sinx+cosx/1+sinx
    we must prove that it is equal
    2secx=cosx(1-sinx)+cosx(1+sinx)/1-sinx+1+sinx

    which give me
    cosx-cosxsinx+cosx+cosxsinx/sinx^2
    from this point i am lost have i make mistake ??
    please use parentheses
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  3. #3
    MHF Contributor
    Joined
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    Quote Originally Posted by gregorio View Post
    we have 2 secx= cosx/1-sinx+cosx/1+sinx
    we must prove that it is equal
    2secx=cosx(1-sinx)+cosx(1+sinx)/1-sinx+1+sinx

    which give me
    cosx-cosxsinx+cosx+cosxsinx/sinx^2
    from this point i am lost have i make mistake ??
    When you combined the two fractions into only one fraction, the denominator should have been (1-sinX)(1+sinX). Not your 1-sinX+1+sinX.
    Like,
    1/2 +1/3 = (1*3 +1*2)/(2*3) = 5/6
    Not: = (1*3 +1*2)/(2+3) = 5/5 = 1. -------No.

    So your second line should have been:
    2secx = cosx(1-sinx) +cosx(1+sinx) / (1-sinx)(1+sinx)

    Let me finish it.
    2secX = cosX(1-sinX +1+sinX) / 1 -sin^2(X)
    2secX = cosX(2) / cos^2(X)
    2secX = 2 / cosX
    2secX = 2(1/cosX)
    2secX = 2secX
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