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?
 A: 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 $x>0$. $\blacksquare$
A: 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.
A: 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, $$2 \arctan{\sqrt{x}} = 2 \theta = \arcsin \left(\frac{x-1}{x+1} \right) + \frac{\pi}{2}$$
A: 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:
$$
2\cdot\frac{\left(\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)}{\sqrt{1+\left(\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)^2}}\cdot \frac{1}{\sqrt{1+\left(\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)^2}} = \frac{1-x}{1+x}.
$$
