Challenging Hyperbolic Proof:

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Challenging Hyperbolic Proof:

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Prove that \[\frac{ 1+tanhx }{ 1-tanhx }=e ^{2x}\]
Medal and fans for first. Already know it.
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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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
You do not need (e^x-e^-x)/(e^x+e^-x) to solve at all. To prove it, it must be gone.
I mean to prove it faster without tedious algebra, it should be gone.
\[\frac{\cosh x }{\cosh x}\frac{ 1+\tanh x }{ 1-\tanh x }=\frac{\cosh x + \sinh x}{\cosh x - \sinh x} = \frac{e^x}{e^{-x}}\]
You missed a between the first and second, but that seems reasonable.
a step*
What step? I skipped several steps but they were simple so no point in wasting time typing :P
If you want me to elaborate on my reasoning anywhere I can explain myself. :D
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Just to clarify. Its all good.
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.
yea and that fact that sinhx+coshx=e^x while coshx-sinhx=e^-x

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