Show an $\arctan$ and $\arcsin$ function is constant Show that for every $x\geq1$ the following is true:
$2\arctan x + \arcsin \frac{2x}{1+x^2} = \pi$
One way (mentioned in the link at the bottom) would be to calculate the derivative of the left side, show that it is always $0$ then show that for $x=1$ the equation is true.
I'm trying for some time to find a cleaner way to prove the equality, without so much algebra. Does anyone have any idea where to start?
Someone already mentioned the same problem here.


\begin{align} 2\arctan(x)+\arcsin\left(\frac{2x}{1+x^2}\right)
&=\pi
\end{align}

Let's $\chi=\arctan(x)\ge\pi/4$ since $x\ge1$.
Consider a unit circle

Then in a $\triangle OAB$ we have $|OA|=|OB|=1$,
$\angle OAB=\angle OBA=\chi$, 
$\angle AOB=\alpha=\pi-2\chi$, 
$|AH|=\sin(\alpha)=a$.
In the $\triangle ABH$
\begin{align}
|AH|&=|BH|\tan(\chi)
\\
\sin(\alpha)
&=
(1-\cos(\alpha))\tan(\chi)
\\
\sin(\alpha)
&=
\left(1-\sqrt{1-\sin^2(\alpha)}\right)\tan(\chi)
\\
a&=
\left(1-\sqrt{1-a^2}\right)x
\\
\left(1-\frac{a}{x}\right)^2&=
1-a^2
\\
\frac{a^2}{x^2}+a^2 &= 2\frac{a}{x}
\\
a\left(\frac{1}{x^2}+1\right) 
&= \frac{2}{x}
\\
a&=\sin(\alpha)=
\frac{2x}{1+x^2}.
\end{align}
