# Hyperbolic function proof

1. Dec 2, 2003

### bard

I need help proving this hyperbolic function

Prove that

$$\tan^{-1}\hbar {x}=\frac{1}{2}\ln\frac{1+x}{1-x}$$

my work
x=e^y-e^-y/e^y+e^-y
(e^y+e^-y)x=e^y-e^-y

0=e^y-e^-y-xe^y+xe^-y

e^y(e^y-e^-y-xe^y+xe^-y)

e^2y-x(e^2y)-1+x=0

I know i have to use the quadratic equation here

Last edited: Dec 2, 2003
2. Dec 3, 2003

### HallsofIvy

Staff Emeritus
One problem you have is that you have "lost" a sign:

instead of e^2y-x(e^2y)-1+x=0, the correct equation is
e^(2y)- x(e^(2y))-1-x= 0 or, more simply, e^(2y)(1-x)= 1+x
so e^(2y)= (1+x)/(1-x).
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