Question: The parametric curve of a lemniscate is given by $\displaystyle \boldsymbol{r} = \left(\dfrac{a\cos t}{1 + \sin^2 t}, \dfrac{a \sin t \cos t}{1 + \sin^2 t}\right)$. Show that the signed curvature is given by $\tilde{\kappa}(t) = \dfrac{-\cos t}{2\sqrt{1+\sin^2t}}.$
My Attempt: I first began by finding the 'speed' of this, which worked out nicely to $\dfrac{|a|}{\sqrt{1+\sin^2 t}},$ however getting there was an absolute nightmare. All other routes seem to be an unforgiving algebra bash involving $\dot{\boldsymbol{r}}$ and $\ddot{\boldsymbol{r}}$ which are extremely messy expressions, so is there something I'm missing? Is there is a simple way to work this out? If not, what's the best approach?
It is not a nightmare if you simplify things one at the time and use multiple angles
$$x=\frac{a \cos (t)}{\sin ^2(t)+1}$$ $$ x'=-\frac{2 a \sin (t) (\cos (2 t)+5)}{(\cos (2 t)-3)^2} \qquad x''=\frac{a (2 \cos (t)+45 \cos (3 t)+\cos (5 t))}{2 (\cos (2 t)-3)^3}$$ Now, for $y$ use the fact that $$y= a \sin(t)\, x$$ $$y'=a\left(\sin (t) x'+ \cos (t)x\right) \qquad y''=a\left(\sin (t) \left(x''-x\right)+2 \cos (t) x'\right)$$