anonymous
  • anonymous
Challenging Hyperbolic Proof:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Prove that \[\frac{ 1+tanhx }{ 1-tanhx }=e ^{2x}\]
anonymous
  • anonymous
Medal and fans for first. Already know it.
anonymous
  • anonymous
I think it's just tedious algebra. Using the definition of tanh(x) = {e^x - e^(-x)] / (e^x + e^-x), substitute back into the equation and simplify

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anonymous
  • anonymous
look here: https://www.wolframalpha.com/input/?i=%5B1+%2B+%28e%5Ex+-+e%5E%28-x%29%29+%2F+%28e%5Ex+%2B+e%5E-x%29%5D+%2F+%5B1+-+%28e%5Ex+-+e%5E%28-x%29%29+%2F+%28e%5Ex+%2B+e%5E-x%29%5D
anonymous
  • anonymous
You do not need (e^x-e^-x)/(e^x+e^-x) to solve at all. To prove it, it must be gone.
anonymous
  • anonymous
I mean to prove it faster without tedious algebra, it should be gone.
Empty
  • Empty
\[\frac{\cosh x }{\cosh x}\frac{ 1+\tanh x }{ 1-\tanh x }=\frac{\cosh x + \sinh x}{\cosh x - \sinh x} = \frac{e^x}{e^{-x}}\]
anonymous
  • anonymous
You missed a between the first and second, but that seems reasonable.
anonymous
  • anonymous
a step*
Empty
  • Empty
What step? I skipped several steps but they were simple so no point in wasting time typing :P
Empty
  • Empty
If you want me to elaborate on my reasoning anywhere I can explain myself. :D
anonymous
  • anonymous
|dw:1439616015632:dw|
anonymous
  • anonymous
Just to clarify. Its all good.
Empty
  • Empty
I think the bigger jump is this one: \[\cosh x - \sinh x = e^{-x}\] since it's kind of subtle to see how this is true. \[\cosh x = \cosh -x\] and \[-\sinh x = \sinh -x\] so we can plug these in to get it.
anonymous
  • anonymous
yea and that fact that sinhx+coshx=e^x while coshx-sinhx=e^-x

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