Solving for $y$ with $\arctan$ I know this is a very low level question, but I honestly can't remember how this is done. I want to solve for y with this:
$$
x = 2.0 \cdot \arctan\left(\frac{\sqrt{y}}{\sqrt{1 - y}}\right)
$$
And I thought I could do this:
$$
\frac{\sqrt{y}}{\sqrt{1 - y}} = \tan\left(\frac{x}{2.0}\right)
$$
But it seems like I've done something wrong getting there. Could someone break down the process to get to $y =$ ?
Again, I know this is very basic stuff, but clearly I'm not very good at this.

So, since Henry told me that I wasn't wrong, I continued and got a really simple answer. Thanks!
x = 2 * arctan(sqrt(y)/sqrt(1 - y))
sqrt(y)/sqrt(1 - y) = tan(x/2)
1/(sqrt(1 - y) * sqrt(1/y)) = tan(x/2)
1/tan(x/2) = sqrt(1/y - 1)
1/(tan(x/2))^2 + 1 = 1/y
y = (tan(x/2))^2/((tan(x/2))^2 + 1)

Thanks again to Henry!
EDIT
Followed by:
y = ((1 - cos(x)) / (1 + cos(x))) / (1 + (1 - cos(x))/(1 + cos(x)))
  = (1 - cos(x)) / ((1 + cos(x)) * (1 + (1 - cos(x))/(1 + cos(x))))
  = (1 - cos(x)) / ((1 + cos(x)) + (1 - cos(x)))
  = (1 - cos(x)) / 2

