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?

One way would be to notice that both functions have the same derivative and then find out what the "constant" is by plugging in $x=0$.
Here's another way.  Look at
$$
\frac\pi2 - 2\arctan\sqrt{x} = 2\left( \frac\pi4 - \arctan\sqrt{x} \right) = 2\left( \arctan1-\arctan\sqrt{x} \right).
$$
Now remember the identity for the difference of two arctangents:
$$
\arctan u - \arctan v = \arctan\frac{u-v}{1+uv}.
$$
(This follows from the usual identity for the tangent of a sum.)  The left side above becomes
$$
2\arctan\frac{1-\sqrt{x}}{1+\sqrt{x}}.
$$
The double-angle formula for the sine says $\sin(2u)=2\sin u\cos u$.  Apply that:
$$
\sin\left(2\arctan\frac{1-\sqrt{x}}{1+\sqrt{x}}\right) = 2 \sin\left(\arctan\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)\cos\left(\arctan\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)
$$
Now remember that $\sin(\arctan u) = \dfrac{u}{\sqrt{1+u^2}}$ and $\cos(\arctan u) = \dfrac{1}{\sqrt{1+u^2}}$
Then use algebra:
$$
2\cdot\frac{\left(\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)}{\sqrt{1+\left(\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)^2}}\cdot \frac{1}{\sqrt{1+\left(\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)^2}} = \frac{1-x}{1+x}.
$$
