Question

Challenging Hyperbolic Proof:

Answer 1

Prove that \[\frac{ 1+tanhx }{ 1-tanhx }=e ^{2x}\]

Answer 2

Medal and fans for first.
Already know it.

Answer 3

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

Answer 4

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

Answer 5

You do not need (e^x-e^-x)/(e^x+e^-x) to solve at all.
To prove it, it must be gone.

Answer 6

I mean to prove it faster without tedious algebra, it should be gone.

Answer 7

\[\frac{\cosh x }{\cosh x}\frac{ 1+\tanh x }{ 1-\tanh x }=\frac{\cosh x + \sinh x}{\cosh x - \sinh x} = \frac{e^x}{e^{-x}}\]

Answer 8

You missed a between the first and second, but that seems reasonable.

Answer 9

a step*

Answer 10

What step? I skipped several steps but they were simple so no point in wasting time typing :P

Answer 11

If you want me to elaborate on my reasoning anywhere I can explain myself. :D

Answer 12

Just to clarify.
Its all good.

Answer 13

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.

Answer 14

yea and that fact that sinhx+coshx=e^x
while coshx-sinhx=e^-x