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-   -   Trig identities (https://www.askmehelpdesk.com/showthread.php?t=286880)

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

    tanx/cscx=secx-cosx
  • Dec 1, 2008, 08:04 AM
    ebaines

    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?
  • Dec 1, 2008, 02:46 PM
    hollyk

    Yes thanks

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