How to prove that $2 \arctan\sqrt{x} = \arcsin \frac{x-1}{x+1} + \frac{\pi}{2}$ I want to prove that$$2 \arctan\sqrt{x} = \arcsin \frac{x-1}{x+1} + \frac{\pi}{2}, x\geq 0$$
I have started from the arcsin part and I tried to end to the arctan one but I failed. 
Can anyone help me solve it?

It's kind of fun to "unveil" these mysterious formulas using only trigonometry, by which they all appear to be very simple angle relationships.
For $x>1$ consider the Figure below.
$\hskip1.5in$ 
Start with a right-angled triangle with hypothenuse $\overline{AC} = x+1$ and side $\overline{BC}=x-1$, so that
$$\alpha = \arcsin\left(\frac{x-1}{x+1}\right),$$
and, by Pythagorean Theorem, 
$$\overline{AB} = 2\sqrt x.$$
Extend $BC$ to a segment $\overline{CD} = x+1.$ Then $\overline{BD} = 2$ and
$$\beta = \arctan \sqrt x.$$
Now use the fact that $\triangle ACD$ is isosceles and $\triangle ABD$ is right-angled to write
$$ \beta + (\beta - \alpha) = \frac{\pi}{2},$$
i.e.
$$2\arctan\sqrt x = \arcsin\left(\frac{x-1}{x+1}\right) + \frac{\pi}{2}.$$
For $0<x<1$ use the Figure below.
$\hskip1.5in$ 
Here $\overline{AC} = x+1$ and $\overline{BC} = 1-x$, so that
$$\alpha = -\arcsin\left(\frac{x-1}{x+1}\right).$$
Again we have $\overline{AB}=2\sqrt x$.
Extend $BC$ to a segment $\overline{BD} = 2$, so that $\overline{AC} = \overline{CD} = x+1$ and
$$\beta = \arctan\sqrt{x}.$$
Since $\triangle ACD$ is isosceles and $\triangle ABD$ is right-angled, we have, this time
$$(\alpha + \beta) + \beta = \frac{\pi}{2}.$$
Once the replacemente is done, this yields again the desired relationship, which therefore is valid for $x>0$. $\blacksquare$
