\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.