How do you simplify \cos(\arctan(x))?

Explanation:
Let ${\displaystyle a=\arctan \left(x\right)}$
The principal value of ${\displaystyle a\in (-\frac{\pi }{2},\frac{\pi }{2})}$
Then ${\displaystyle \tan \alpha =x}$ and
${\displaystyle 0\le \cos \alpha \in [0,1)}$ (wrongly marked as [0,-1], in my previous answer, two years ago).
Now the given expression is
${\displaystyle \cos \alpha =\frac{1}{\sqrt{1+x^2}},x\in (-\frac{\pi }{2},\frac{\pi }{2})}$
It is important that ${\displaystyle \cos \alpha \ge 0}$, for ${\displaystyle \alpha \in Q_1}$ or ${\displaystyle Q_4}$
If the piecewise-wholesome general inverse operator
${\displaystyle \left({\tan }^{-1}\right)}$ is used
${\displaystyle {\cos \left(\tan \right)}^{-1}x,=\pm \frac{1}{\sqrt{1+x^2}}}$
the negative sign is chosen, when ${\displaystyle x\in Q_3}$
Example:
${\displaystyle \cos \left(\arctan 1\right)=\frac{1}{\sqrt{2}},\arctan 1=\frac{\pi }{4}}$
${\displaystyle {\cos \left(\tan \right)}^{-1}=\cos (k\pi +\frac{\pi }{4},k=0,\pm 1,\pm 2,\pm ,\dots }$
${\displaystyle \in Q_1}$ or ${\displaystyle Q_3,x=\dots \frac{\pi }{4},\frac{5}{4}\pi ,\dots }$
So, the value is ${\displaystyle \pm \frac{1}{\sqrt{2}}}$
My intention, in this approach, is to inform about nuances.