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    hollyk's Avatar
    hollyk Posts: 2, Reputation: 1
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    #1

    Nov 30, 2008, 09:39 PM
    trig identities
    I can't seem to figure this out...

    tanx/cscx=secx-cosx
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #2

    Dec 1, 2008, 08:04 AM

    It almost always works best if you substitute sines and cosines for the tangent, cosecant etc.

    The left hand side is:

    tanx/cscx = (sinx/cosx)*(sinx/1) = sin^2x/cosx

    The right hand side is:

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

    Can you take it from here?
    hollyk's Avatar
    hollyk Posts: 2, Reputation: 1
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    #3

    Dec 1, 2008, 02:46 PM

    Yes thanks
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