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    How do I do...

    (1 - cosx)(1+secx) = sinxtanx

    Should I expand the left hand side? This is what I have done...but I still can't get it so that it's like the RHS where sinx x sinx/cosx

    Thank you
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    Where did you get the equation from? It's not a trig identity that holds for all x.
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    (Original post by ttoby)
    Where did you get the equation from? It's not a trig identity that holds for all x.
    Whoops, mistake. It's + secx not - secx. :o:

    And from the edexcel c3 textbook
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    Expand the LHS, should come out pretty easily.
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    This is how I did it...

    sin(x)tan(x)= sin^2(x)/cos(x)= (1-cos^2(x))/cos(x)= 1/cosx - cosx = secx - cosx = LHS

    Hopefully that helps.
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    ok doing it nw
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    (Original post by insparato)
    Expand the LHS, should come out pretty easily.
    Got it! :woo:

    Thanks for your help.

    Solved.
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    just start on the left hand side expand the brackets then simplify.
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    When i did it i got down to secx - cosx but whats the last and final step to get it into sinxtanx?
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     \frac{1}{cosx} - cosx Combine that.
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    expand the brackets (LHS):
    =1+secx-cosx-(cosx)(secx)
    =1+(1/cosx)-cosx-cosx(1/cosx)
    =1+(1/cosx)-cosx-1
    =(1/cosx)-cosx
    =(1/cosx)-(cosx/1)
    =(1-cos^2x)/cosx <--cross multiply to get same denominator
    =sin^2x/cosx <--use of identity sin^2x +cos^2x =1
    =tanxsinx


    gd luck wid revision + exam on wednesday
 
 
 
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