## Differential geometry question

$\sigma(u,v)=(u-sinucoshv,1-cosucoshv,-4sin\frac{u}{2}sinh\frac{v}{2}$

a)show that this is a conformally parametrised surface.

According to the definition in my book, this is the case when from the first fundamental form:
$Edu^2$ $+2Fdudv$ $+Gdv^2$
we have:

$E=$ $\sigma_u\cdot\sigma_u$ $=G=$ $\sigma_v\cdot\sigma_v$ and
$F=$ $\sigma_u\cdot\sigma_v=0$

I get the following:
$F= 0$
$E=$ $1-2cosucoshv+cosh^2v+4cos^2\frac{u}{2}sinh^2\frac{v} {2}$
$G=$ $sinh^2v$ $+4sin^2\frac{u}{2}cosh^2\frac{v}{2}$

The book has the following:
$F=0$
$E=G=(coshv+1)(coshv-cosu)$

I have tried every trigonometric and hyperbolic identity but i cant get this solution.