How to prove that $2 \arctan\sqrt{x} = \arcsin \frac{x-1}{x+1} + \frac{\pi}{2}$

Define $\displaystyle g(x)=2\arctan(\sqrt{x})-\arcsin\left(\frac{x-1}{x+1}\right)-\frac{\pi}{2}$. Differentiate to find that $g'(x)=0$. Since this was true on the connected set $[0,\infty)$ you can conclude that $g$ is contant. Note then that $g(0)=0$ to finish.

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, $$2 \arctan{\sqrt{x}} = 2 \theta = \arcsin \left(\frac{x-1}{x+1} \right) + \frac{\pi}{2}$$

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.

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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.

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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 $x>0$. $\blacksquare$