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Roadblock in trig identity

  1. Oct 13, 2006 #1

    I am trying to prove [tex]\frac{tanx+1}{tanx-1}=\frac{secx+cscx}{secx-cscx}[/tex]

    I am working on the right side and trying to get it to the same form as the left side.

    I get to [tex]\frac{1+2sinxcosx}{sin^{2}x-cos^{2}x}[/tex] which I COULD apply the fact that sin2x=2sinxcosx as well as cos^2x-sin^2x=cos2x (after multiplying both top and bottom by -1) but that gets me to

    [tex]\frac{-Sin(2x)-1}{cos(2x)}[/tex] which looks completely useless!

    Anyone see something I don't? Just point me in the right direction :) Thanks again :biggrin:
  2. jcsd
  3. Oct 13, 2006 #2
    oh my god why didn't I see that! lol. I was looking at that step ( I had the exact same work shown) and thought that was an important step, but dividing by cos x just never popped into my head, lol. Anyways, thanks a ton :).
  4. Oct 13, 2006 #3


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    Staff Emeritus
    Science Advisor

    Actually, starting from
    and multiplying the numerator and denominator of the right side by sin x gives the result immediately.
  5. Oct 13, 2006 #4
    lol, how do you see this stuff? I don't even start thinking about solving it until at least the 3rd step. Maybe I answered my own question here... lol :)

    Thanks to both of you, I am going to start looking for a way to immediately put it into the desired form at EVERY step from now on.
    Last edited: Oct 13, 2006
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