Tricky hyperbolic function proof.

i) Show that (coth(x)+1)/(coth(x)-1) = e^(2x)


So,

coth(x) = cosh(x)/sinh(x)

cosh(x) = (e^(x)+e^(-x))/2

and

sinh(x) = (e^(x)-e^(-x))/2

Thus:

coth(x) = (e^(x)+e^(-x))/(e^(x)-e^(-x)) ------- A


I then replace the coth(x) in the original with A. But I can't seem to do anything with it..

I have been trying for 2 hours I've tried multiplying by all sorts of 1s with the intention of making things simpler.

Can anyone point me in the right direction?

Hint: Multiply your expression for coth(x) by $\frac{e^{x}}{e^{x}}$.

I had already tried that.. I got to the following expression:

Coth(x) = (e^2x+1)/(e^2x-1))

So the original becomes:

[((e^2x+1)/(e^2x-1))+1] / [((e^2x+1)/(e^2x-1)-1)]

But I dismissed it as I thought I'd taken a wrong turn.

What next?

I think my brain has melted to the point where I am missing something obvious..

Ahh I got it.. Thank you kindly

+ rep