<s>[INST] 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 $$a=\frac{1-t^2}{1+t^2} \quad \text{and} \quad b=\frac{2t}{1+t^2}$$

 [/INST] Since $x^2 + y^2 = 1$ draws a circle on the Euclidean plane, we have $(x, y) = (\cos \theta , \sin \theta)$ for some $\theta$. Now let $\tan (\theta/2) = t$, then $\tan \theta = \dfrac{2t}{1-t^2}$, $\cos\theta = \dfrac{1-t^2}{1+t^2}$, $\sin\theta = \dfrac{2t}{1+t^2}$ by trigonometry. 
Another way to see this : 
Let $(0, u)$ be a point on $y$-axis, and consider a line joining $(0, u)$ and $(1, 0)$. This line and the circle meet twice: once at $(1, 0)$, and by calculation the another point is at $\left(\dfrac{ u^2 - 1}{u^2 + 1}, \dfrac{2u}{ u^2 + 1}\right)$. Consider $t = 1/u$ (with $u\ne0$) then you have the wanted parametrization. You can recover the case of $t = 0$ and $t = \infty$ easily.
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