Evaluate a rational function of $x,y,z$ given two polynomial equations in $x,y,z$ Let $x, y, z$ be real numbers. Given that $$2x(y^2−1)+2y(x^2−1)=(1+x^2)(1+y^2)$$ and $$4z(1−y^2)+4y(1−z^2)=(1+z^2)(1+y^2)$$ Find the value of the following expression:
$$\Bigg(\frac{2x}{1+x^2}−\frac{2z}{1+z^2}\Bigg)^2+\Bigg(\frac{1−z^2}{1+z^2}−\frac{1−x^2}{1+x^2}\Bigg)^2$$ How do I use $x=\tan\alpha$ and $z=\tan\beta$ in this question. I know it will be useful as the expression to be found out is screaming trigonometric substitution. These were the two equations I finally got after substitution-$$(y^2-1)\sin2\alpha-2y\cos2\alpha=1+y^2$$ and $$2(1-y^2)\sin2\beta+4y\cos2\beta=1+y^2$$ and the expression to be found is $-2(\sin2\alpha\sin2\beta-2\cos2\alpha\cos2\beta)$. Now how do I manipulate my equations? Thanks.

I don't know if a better solution is possible, but here is mine.Consider the following points:
$A(\frac{2x}{1+x^2},\frac{1−x^2}{1+x^2}),B(\frac{1−y^2}{1+y^2},\frac{2y}{1+y^2}),C(\frac{2z}{1+z^2},\frac{1−z^2}{1+z^2})$.
Notice that these points on a unit circle, as for any real number $a$,
$$(\frac{2a}{1+a^2})^2+(\frac{1−a^2}{1+a^2})^2=1$$
According to the assumptions of the problem, we have that
$$AB^2=(\frac{2x}{1+x^2}−\frac{1−y^2}{1+y^2})^2+(\frac{1−x^2}{1+x^2}−\frac{2y}{1+y^2})^2$$$$=2−2⋅\frac{2x(1−y^2)+2y(1−x^2)}{(1+x^2)(1+y^2)}=4$$
Therefore, $AB=2$. On the other hand, we have that
$$BC^2=(\frac{2y}{1+y^2}−\frac{1−z^2}{1+z^2})^2+(\frac{1−y^2}{1+y^2}−\frac{2z}{1+z^2})^2$$$$=2−2⋅\frac{2z(1−y^2)+2y(1−z^2)}{(1+z^2)(1+y^2)}=1$$
Hence, we obtain that $BC=1$. From the condition $AB=2$, it follows that $AB$ is a diameter, thus $∠C=90°$ and $AC^2=3$. Hence,
$$(\frac{2x}{1+x^2}−\frac{2z}{1+z^2})^2+(\frac{1−z^2}{1+z^2}−\frac{1−x^2}{1+x^2})^2=AC^2=3.  $$ 
