## anonymous one year ago Challenging Hyperbolic Proof:

1. anonymous

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

2. anonymous

Medal and fans for first. Already know it.

3. 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

4. anonymous
5. anonymous

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

6. anonymous

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

7. Empty

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

8. anonymous

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

9. anonymous

a step*

10. Empty

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

11. Empty

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

12. anonymous

|dw:1439616015632:dw|

13. anonymous

Just to clarify. Its all good.

14. 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.

15. anonymous

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