How can I prove this trigonometric equality? $$\arccos\left(\frac{1-x^2}{1+x^2}\right) = 2\arctan{x}$$ for $x \geq 0$.
I'm not even sure what kind of math to try? :(

Put $t=\arctan x$. Then $x=\tan t$.
You have to prove that $2t=\arccos(1-x^2)/(1+x^2)$.
This is more-or-less the same as $(1-x^2)/(1+x^2)=\cos2t$. If you put
$x=\tan t$ into $(1-x^2)/(1+x^2)$ what do you get?
