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    Prove that:

    cot2x + cosec2x = cotx

    This is what I've done so far, which gives me the wrong answere.

    cot2x +cosec2x

    (1-tan^2x)/(2tanx) + 1/(2sinxcosx)

    [(1-sin^2x/cos^2x) / (2sinx/cosx)] + 1/(2sinxcosx)

    (cosx)/(2sinx) + 1/(2sinxcosx)

    (cosx + 1)/(2sinx)

    (cotx + 1)/2

    Please Help.
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    [(1-sin^2x/cos^2x) / (2sinx/cosx)]
    =(cosx - (sin^2x/cosx)) / 2sinx

    the other fraction shares 2sinx, so add them together and times this one by cos x

    along those lines anyway.

    what you wrote for that fraction isnt right

    it comes to cos^2x - sin^2x + 1 / (2sinxcosx)
    which comes out to cot x = cosx/sinx
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    I've included that though, what do you mean?

    (to stevo)
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    cot2x = 1/tan2x
    cosec2x = 1/sin2x

    cos2x/sin2x + 1/sin2x = [cos2x + 1]/sin2x

    cos2x = (cos^2)x - (sin^2)x therefore cos2x + 1 = (cos^2)x - (sin^2)x + 1
    sin2x = 2sinxcosx

    [cos2x+1]/[sin2x] = [(cos^2)x - {(sin^2)x - 1}]/[2sinxcosx]
    = [(cos^2)x - {-(cos^2)x}]/[2sinxcosx]
    = [2(cos^2)x]/[2sinxcosx]
    = (cos^2)x / sinxcosx
    = cosx / sinx
    = cotx

    edit: decided to show whole method...
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    OHHHH!

    Thank you. I see it now, that's really annoying.
 
 
 
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Updated: January 13, 2005

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