Trigonometric equation inversion I am trying to invert the following equation to have it with $\theta$ as the subject:
$y = \cos \theta \sin \theta - \cos \theta -\sin \theta$
I tried both standard trig as well as trying to reformulate it as a differential equation (albeit I might have chosen an awkward substitution). Nothing seems to stick and I keep on ending up with pretty nasty expressions. Any ideas?

Rewrite the equation as $(1-\cos\theta)(1-\sin\theta)=1+y.$
Now make the Weierstrass substitution $t=\tan(\theta/2)$. It is standard that 
$\cos\theta=\frac{1-t^2}{1+t^2}$ and $\sin\theta=\frac{2t}{1+t^2}$. So our equation becomes 
$$\frac{2t^2}{1+t^2}\cdot \frac{(1-t)^2}{1+t^2}=1+y.$$
Take the square root, and clear denominators. We get
$$\sqrt{2}t(1-t)=\sqrt{1+y}(1+t^2).$$
This is a quadratic in $t$. 
