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One more trig identity

  1. May 5, 2009 #1
    1. The problem statement, all variables and given/known data
    I'm finally starting to understand proving trig identities, but I have just one more that I can't seem to figure out.

    secx - tanxsinx = cosx


    2. Relevant equations
    N/A


    3. The attempt at a solution
    Well first, I multiplied the tanx and sinx and came up with sin2x / cosx
    Now I'm stuck. I'm trying to do 1/cosx - sin2x / cosx
    Would the GCF be (cosx)(cosx) or just cosx, I'm drawing a blank here. I'm debating whether or not to change the sin2x to 1 - cos2x, but even if I do, I can't figure out how that would simplify to just cosx
     
  2. jcsd
  3. May 5, 2009 #2

    Cyosis

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    You're almost there[tex]\frac{1}{\cos x}-\frac{\sin^2x}{\cos x}=\frac{1}{\cos x}(1-\sin^2x)[/tex]. You should notice something very familiar now.
     
  4. May 5, 2009 #3
    So would it simplify to cos2x / cosx, which would then simplify to cosx for my answer?
     
  5. May 5, 2009 #4

    Cyosis

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    Yep it's that easy!
     
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