If $a^2 + b^2 = 1$, show there is $t$ such that $a = \frac{1 - t^2}{1 + t^2}$ and $b = \frac{2t}{1 + t^2}$ My question is how we can prove the following:

If $a^2+b^2=1$, then there is $t$ such that $$a=\frac{1-t^2}{1+t^2} \quad \text{and} \quad b=\frac{2t}{1+t^2}$$


As its an equation and open end it will have infinite solutions so Assume $x=\sin(2\theta)$, $y=\cos(2\theta)$ as we all know that sum of their squares is $1$ now by double angle $\sin(2\theta)=\frac{2t}{1+t^2}$, $\cos(2\theta)=\frac{1-t^2}{1+t^2}$ and now you can prove it trigonometrically. 
