Proof of trig identity using t-formulae Show that $$2\arctan x = \arccos\frac{1-x^2}{1+x^2}$$
if $ x > 0$, and 
$$2\arctan x = -\arccos\frac{1-x^2}{1+x^2}$$ if $x <0.$
I have been able to equate the expressions by letting $x=\tan(y/2)$, but I do not know how to show the different signs of the equation for $x>0$ and $x<0$.

The simplest way to go is probably to calculate $\;\cos(2\arctan x)$. 
Set $\theta=\arctan x$ and use the duplication formula:
\begin{align}
\cos 2\theta&=\frac{1-\tan^2\theta}{1+\tan^2\theta}=\frac{1-x^2}{1+x^2},\\
\text{so }\qquad
2\arctan x&\equiv \pm\arccos\frac{1-x^2}{1+x^2}\mod 2\pi.
\end{align}
Now either $\;0\le \theta<\frac\pi2,\;$  so $\;0\le 2\theta<\pi$, which is (included in, strictly speaking) the range of  arccos. In this case we conclude that
$$ 2\arctan x=\arccos\frac{1-x^2}{1+x^2},.$$
or $\;-\frac\pi 2<\theta<0$, so $\;-\pi<2\theta<0$, and for the same reason, we conclude in this case that
$$ 2\arctan x=-\arccos\frac{1-x^2}{1+x^2}.$$
