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?

Let $\arctan{\sqrt{x}} = \theta$. Then we have $x = \tan^2 (\theta)$. Hence, $$\frac{x-1}{x+1} = \frac{\tan^2 (\theta)-1}{\tan^2 (\theta)+1} = \sin^2(\theta) - \cos^2(\theta) = - \cos(2 \theta)= \sin \left(2 \theta - \frac{\pi}{2} \right)$$
Hence, $$2 \arctan{\sqrt{x}} = 2 \theta = \arcsin \left(\frac{x-1}{x+1} \right) + \frac{\pi}{2}$$
