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Sech^2(x) = 1 - tanh^2(x) proof - need help!

  1. Jun 4, 2012 #1


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    1. The problem statement, all variables and given/known data

    Prove Sech^2(x) = 1 - tanh^2(x)

    2. Relevant equations

    TanH(x) = (e^x - e^-x)/(e^x+e^-x)
    CosH(x) = (e^x+e^-x)/2
    SinH(x) = (e^x - e^-x)/2

    3. The attempt at a solution

    SecH^2(x) = 1/cosh^2(x)
    =1 / (e^x - e^-x)^2 / 4
    =4/(e^x - e^-x)^2

    This is where I am stuck. Any help is greatly appreciated. Thank you !
  2. jcsd
  3. Jun 4, 2012 #2
    I'd recommend proving a simpler statement which directly leads to the statement you want to prove.

    Try proving Cosh^2(x)-sinh^2(x)=1
  4. Jun 4, 2012 #3
    Try working from the more complicated side and work towards the simpler side. Often when you do this, terms cancel somewhere. If you start from the simpler side you usually need to creatively add 0 or multiply by 1, and this is often not that easy to see. Simplifying is often easier to see.
  5. Jun 4, 2012 #4


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    Use you expression for tanh(x). Square that, then use a common denominator to combine 1 - tanh2(x) into one fraction.
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