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 $$a=\frac{1-t^2}{1+t^2} \quad \text{and} \quad b=\frac{2t}{1+t^2}$$


We show that except for the case $a=-1$, $b=0$, there always is a $t$ satisfying the conditions of the OP. The calculation is geometric. A similar calculation is of importance at the beginning of the theory of elliptic curves.
Let $P=(a,b)\ne(-1,0)$ be on the unit circle $x^2+y^2=1$. Suppose that the line joining $(-1,0)$ to $(a,b)$ has slope $t$. The line has equation $y=t(x+1)$. Substituting in $x^2+y^2=1$ we get after a little calculation that $(1+t^2)x^2+2tx+t^2-1=0$. One of the roots is $-1$. The product of the roots is $\frac{t^2-1}{1+t^2}$, so the other root $a$ is given by
$$a=\frac{1-t^2}{1+t^2}.$$
Now from $y=t(x+1)$ we get 
$$b=t\left(\frac{1-t^2}{1+t^2}+1\right)  =\frac{2t}{1+t^2}.$$
Remark: The "exceptional" point $(-1,0)$ can be dealt with by letting $t=\infty$.
