Why is $[-\frac{\pi}{2}, \frac{\pi}{2} ]$ the set of values for $f(x) = \arctan \sqrt{x^2-1} + \arcsin \frac{1}{x}$? I am given the function:
$$f:D \rightarrow \mathbb{R} \hspace{3cm} f(x) = \arctan \sqrt{x^2-1} + \arcsin \frac{1}{x}$$
where $D$ is the maximum domain of the function. I am told that the set of values of the function is $\bigg [ -\dfrac{\pi}{2} 
,\dfrac{\pi}{2} \bigg ]$. How was this answer reached? I assume derivatives and limits have been used, but I am not sure. If you could show me the steps taken to reach this conclusion, or even just tell me what I need to do, I'd appreciate it.

Hint
Let $\arctan\sqrt{x^2-1}=y,\dfrac\pi2>y\ge0,x=\pm\sec y$
If $x>0,x=\sec y$
$\arcsin\dfrac1x=\arcsin(\cos y)=\dfrac\pi2-\arccos(\cos y)=\dfrac\pi2-y$
If $x<0,x=-\sec y$
$\arcsin(-\cos y)=-\arcsin(\cos y)=?$
