How Can You Prove This Hyperbolic Function Identity?

[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

 

[HallsofIvy]

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).

 

[M98Ranger]

, but i am not sure how to proceed with the proof.

To prove this hyperbolic function, we can use the definition of the tangent inverse function, which states that \tan^{-1}x=y if and only if \tan y=x. Therefore, we need to show that \tan \left(\frac{1}{2}\ln\frac{1+x}{1-x}\right)=x.

Using the double angle formula for tangent, we have:

\tan \left(\frac{1}{2}\ln\frac{1+x}{1-x}\right)=\frac{2\tan\left(\frac{1}{2}\ln\frac{1+x}{1-x}\right)}{1-\tan^2\left(\frac{1}{2}\ln\frac{1+x}{1-x}\right)}

Now, let's substitute the value of x=e^y-e^-y/e^y+e^-y that we have obtained from the given equation:

\tan \left(\frac{1}{2}\ln\frac{1+x}{1-x}\right)=\frac{2\tan\left(\frac{1}{2}\ln\frac{1+e^y-e^-y/e^y+e^-y}{1-e^y+e^-y/e^y+e^-y}\right)}{1-\tan^2\left(\frac{1}{2}\ln\frac{1+e^y-e^-y/e^y+e^-y}{1-e^y+e^-y/e^y+e^-y}\right)}

Now, using the identity \tan\left(\frac{1}{2}y\right)=\frac{\sin y}{1+\cos y}, we can simplify the expression further:

\tan \left(\frac{1}{2}\ln\frac{1+x}{1-x}\right)=\frac{2\left(\frac{\sin\left(\frac{1}{2}\ln\frac{1+e^y-e^-y/e^y+e^-y}{1-e^y+e^-y/e^y+e^-y}\right)}{1+\cos\left(\frac{1}{2}\ln\frac{1+e^y-e^-y/e^y+e^-y}{1-e^y+e^-y/e^y+e^-y}\