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Proving a Trigonometric Identity

  1. Jun 29, 2008 #1
    Im supposed to verify that (1-sinx)/(1+sinx) = (secx-tanx)^2

    RHS = (secx-tanx)^2 = (1/cosx - sinx/cosx)^2 = [(1-sinx) / cosx]^2

    = [(1-sinx)(1-sinx)]/cosx^2 = (1-2sinx+sinx^2)/(1-sinx^2)

    From here, I'm feeling pretty confused. I'm not even sure if all my values are correct.
  2. jcsd
  3. Jun 29, 2008 #2
    From there it's important to recognize that (1-a^2)=(1+a)(1-a) and then (1-2a-a^2)=(1-a)^2=(1-a)(1-a). Then it's just a matter of removing the common term.


    http://www.theUndergraduateJournal.com [Broken]
    Last edited by a moderator: May 3, 2017
  4. Jun 29, 2008 #3

    starting from: [(1-sinx) / cosx]^2 = [(1-sinx)(1-sinx)]/cosx^2

    using pythagorean identity

    [(1-sinx)/(1-sinx)]/(1-sinx^2) = [(1-sinx)/(1-sinx)]/[(1-sinx)(1+sinx)] = (1-sinx)/(1+sinx) = RHS

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