How do I simplify this expression further?

How do I simplify this expression further?

Okay so my adventure starts with this equation $$|x|+|y|=1$$ This makes a
square so what I wanted two do is turn it around so it's sitting "right".
My plan was to convert the equation to polar and then add $\pi/4$ to
$\theta$.



Conversion to polar
It was a straight forward task and I got this $$|r\cos\theta|
+|r\sin\theta|=1$$Now I needed to clean the equation up a little bit so I
collected the $r$ terms as such: $$\begin{align}1&=|r\cos\theta|
+|r\sin\theta|\\&=r|\cos\theta|+r|\sin\theta|\\&=r\left(|\cos\theta|+|\sin\theta|\right)\\1/r&=|\cos\theta|+|\sin\theta|\end{align}$$
Two $\theta$s got messy so I rearranged the equation to have one $\theta$
$$\begin{align}1/r&=|\cos\theta|+|\sin\theta|\\1/r^2&=\cos^2\theta+\sin^2\theta+|2\sin\theta\cos\theta|\\1/r&=\sqrt{1+|\sin2\theta|}\end{align}$$Using
the trig identies $\sin^2\theta+\cos^2\theta=1$ and
$2\sin\theta\cos\theta=\sin2\theta.$



So now that I have my desired polar equation I went about turning the
square. To achieve a $\pi/4$ turn I had to add this to each $\theta$.
Which gets me a new "$\theta$". If you recall the sin function within the
last equation had the $\theta$ so the new operand in that function becomes
$$2\theta\implies2\theta+\pi/2$$ This is the equation that I desire for a
square. It works and I am happy with it. However I need to turn this back
into Cartesian coordinates.



Conversion into Cartesian
Now that I have $$1/r=\sqrt{1+|\sin(2\theta+\pi/2)|}$$ I imideately
realise that there are two trig identities hidden in there. First a sin -
cosine relation. $$\sin(2\theta+\pi/2)=\cos(2\theta)$$ then a double angle
$$\cos(2\theta)=2\cos^2\theta-1$$ Now we have
$$1/r=\sqrt{1+|2\cos^2\theta-1|}$$ It is ready to be converted. So I start
with $r=\sqrt{x^2+y^2}$ and $\theta=\arctan\left(\frac yx\right)$.
$$\frac1{\sqrt{x^2+y^2}}=\sqrt{1+\left|2\cos^2\left[\arctan\left(\frac
yx\right)\right]-1\right|}$$ To see if it is still a square I checked and
it is. So I am assuming that the mistake will come up after.Okay so I am
using trig identities so evaluate $\cos^2\{\arctan(y/x)\}$ $$\begin{align}
\cos^2\left[\arctan\left(\frac yx\right)\right]
=\frac1{y^2/x^2+1}\end{align}$$ Putting it all together
$$\begin{align}\frac1{\sqrt{x^2+y^2}}&=\sqrt{1+\left|\frac2{y^2/x^2+1}-1\right|}\\\frac1{x^2+y^2}&=1+\left|\frac2{y^2/x^2+1}-1\right|\end{align}$$
This is where I get stuck as I have no idea how to collect the $y$ from
the RHS.



I should make it clear that I want a single $y$ term on the LHS and all
the $x$ terms on the RHS.