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Prove 2coth^(-1)(e^x) = ln[cosech x + coth x]

  1. Dec 5, 2011 #1

    sharks

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    Gold Member

    The problem statement, all variables and given/known data
    Prove 2coth^(-1)(e^x) = ln[cosech x + coth x]

    The attempt at a solution
    I'm starting with the L.H.S. first, which is conventional, i think.

    2coth^(-1)(e^x)

    = 2tanh^(-1)(e^-x)

    = 2[1/2 ln [(1+(e^-x))/(1-(e^-x))]

    = ln [(1+(e^-x))/(1-(e^-x))]

    I'm stuck.

    So, i tried to cheat a little, by expanding the R.H.S. and see if i could find a way to convert the last step above into the R.H.S. but my copybook has become something of a painting.
     
  2. jcsd
  3. Dec 5, 2011 #2

    Curious3141

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    Homework Helper

    Hints: work from the RHS, and substitute u = e^x. Will make everything much clearer.
     
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