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? :(

Setting $x=\tan a$ your equality
$$\arccos \dfrac{1-x^{2}}{1+x^{2}}=2\arctan x\qquad (1)$$
becomes
$$\cos 2a=\dfrac{1-\tan ^{2}a}{1+\tan ^{2}a}.\qquad (2)$$
To prove (2) which is listed in the wikipedia we can take the duplication formula
$$\cos 2a=\cos ^{2}a-\sin ^{2}a$$
Dividing by $\cos ^{2}a+\sin ^{2}a=1$, we establish
$$\cos 2a=\dfrac{\cos ^{2}a-\sin ^{2}a}{\cos ^{2}a+\sin ^{2}a}=\dfrac{1-\tan
^{2}a}{1+\tan ^{2}a}.$$
Edit: One must note that $\arctan x$ is an odd function, while $\arccos \frac{1-x^{2}}{1+x^{2}}$ is an even one. Only for $x\geq 0$ are both sides of $(1)$ equal. Identity $(2)$ is valid for both positive and negative values of $a$.

Blue: $y=\arccos((1-x^2)/(1+x^2))$ for $x\ge 0$;  Red: $y=2\arctan x$ 
