simple proof $f(z) = \arcsin(\frac{2z}{(1+z^2)}) + 2\arctan(z)$ $f(z) = \arcsin(\frac{2z}{(1+z^2)}) + 2\arctan(z)$
So the thing that I need to do is to show that $f(2019) = \pi$.
So the thing that I have tried is to calculate $f(0), f(1)$, and to see some kind of connection (or recurrence relation), so I can easily calculate $f(2019)$, but I couldn't find any.
Any tips?
 A: hint
For $z>1$,
$$f'(z)=$$
$$\frac{1}{\sqrt{1-(\frac{2z}{1+z^2})^2}}\frac{2(1+z^2)-4z^2}{(1+z^2)^2}+\frac{2}{1+z^2}=0$$
thus $f$ is constant at $[1,+\infty)$ and
$$f(2019)=\lim_{z\to+\infty}f(z)=0+2\frac{\pi}{2}=\pi$$
or
$$f(2019)=f(1)=$$
$$\arcsin(1)+2\arctan(1)=\pi$$
A: Start by writing 
$$z=tan(\theta)$$
So, for tan$(\theta)=2019$, $\theta$ will lie somewhere in between $\pi /4$ and $\pi/2$.
Now the second term will become 
$$=2arctan(tan(\theta))=2\theta$$
This is allowed as $\theta$ falls within the range of arctan$(x)$
Now taking the first term
$$=arcsin\left(\frac{2tan(\theta)}{1+tan^2(\theta)}\right)$$
$$=arcsin\left(2sin(\theta)cos(\theta)\right)=arcsin\left(sin(2\theta)\right)$$
Since $\pi/4\lt\theta \lt \pi/2$, 
$\pi/2\lt2\theta \lt \pi$
This falls outside the range of arcsin$(x)$
Now by symmetry around $\pi/2$ you can see that for it to come back within the range, for such $x$ we use $\pi - x$
Therefore
$$arcsin\left(sin(2\theta)\right)=\pi - 2\theta$$
Adding the first and second terms together 
$$=\pi - 2\theta + 2\theta = \pi$$
I am attaching a link for more explanation
$\arcsin(\sin x)$ explanation?
