## Differential geometry question

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

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:
$\displaystyle Edu^2$$\displaystyle +2Fdudv$$\displaystyle +Gdv^2$
we have:

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

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

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

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