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.
 A: Outline: We can do it with just trigonometric identities.  Note first that 
$$\tan(2\arctan x)=\frac{2x}{1-x^2}.\tag{1}$$
Then note that 
$$\tan\left(\arcsin\left(\frac{2x}{1+x^2}\right)\right)=\frac{2x}{x^2-1} \tag{2}$$
because $x\ge 1$. So the right-hand side of (2) is the negative of the right-hand side of (1).
No derivatives!
A: We can show $\frac {2x}{1-x^2}=\tan 2 \arctan x = \tan (\pi-\arcsin \frac {2x}{1+x^2}=\tan \arcsin \frac {2x}{1+x^2})$  The left equality comes from $\tan 2y = \frac {2 \tan y}{1-\tan^2 y}$ and the second from drawing a right triangle with one angle $\arcsin \frac {2x}{1+x^2}$  As $\arctan$ is a bijection over this range, we are done.
A: Let $f$ be the function 
$$f(x)=2 \arctan x+\arcsin\left(\frac{2x}{1+x^2}\right)$$
Note that the arcsine function can be expressed alternatively as the arctangent function using $\arcsin y=\arctan \frac{y}{\sqrt{1-y^2}}$ for $|y|<1$.
Here, we have $y= \frac{2x}{1+x^2}$ for $|x|>1$.  Thus, 
$$\arcsin \frac{2x}{1+x^2}=\arctan \frac{2x}{|1-x^2|}=\arctan \frac{2x}{x^2-1}$$
We also note that for $x>1$, $\arctan x>0$.  Therefore, we have 
$$2 \arctan x=\pi+\arctan \frac{2x}{1-x^2}$$
since $\arctan \frac{2x}{1-x^2}<0$. 
Finally, we obtain the coveted result
$2 \arctan x+\arcsin\left(\frac{2x}{1+x^2}\right)=\pi$.
A: This can be solved using basic Inverse Trigonometric Identities
$$2\arctan(x) + \arcsin{\dfrac{2x}{1+x^2}} $$
$$\Rightarrow \arctan\dfrac{2x}{1-x^2} + \arctan\dfrac{2x}{x^2-1}$$
$$\Rightarrow \arctan\dfrac{2x}{1-x^2} + \arctan\dfrac{-2x}{1-x^2}$$
$$\Rightarrow \arctan\dfrac{0}{1+\Big(\dfrac{2x}{1-x^2}\Big)^2}$$

Now since $x\geq 1$
$$ \Rightarrow \arctan(0) = \pi$$


I guess you can also use Andre Nicolas' answer as a hint!
A: if $x\geq 1$, then $x=\tan\frac{\theta}{2}$ for some $\theta\in\left[\frac{\pi}{2},\pi\right)$, then:
$$2\arctan x + \arcsin\frac{2x}{1+x^2} = 2\arctan\tan\frac{\theta}{2}+\arcsin\sin\theta=\theta+(\pi-\theta)=\pi. $$
A: i will use the letter $m$ so that i can interpret it as the slope of the radius in the init circle. let $$t = \tan^{-1}(m), m \ge 1.\tag 1$$  this implies $$\pi/4 \le t < \infty, \tan t = m, \cos t = \frac 1{\sqrt {1+m^2}}, \sin t = \frac m{\sqrt{1+m^2}}. $$  we also know from the double angle formula $$ \sin 2t = 2\sin t \cos t = \frac{2m}{1+m^2}$$ since $\pi/2 \le 2t < p,$ we have $$\sin^{-1}(\sin 2t)) = \pi - 2t.  \tag 2$$
putting $(1)$ and $(2)$ gives $$ \text{ if } 1\le m \text{ then } 2 \tan^{-1}m + \sin^{-1}\left(\frac{2m}{1+m^2}\right)= \pi.$$
A: 
\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}
