Based on the link I shared above, the value of [tex]\arctan(-\sqrt3)[/tex] is that number [tex]x[/tex] that satisfies
[tex]\tan x=-\sqrt3[/tex]
where [tex]x[/tex] lies in what you might call the "principal branch". That is, [tex]\arctan x[/tex] is bounded below by [tex]-\dfrac\pi2[/tex] and above by [tex]\dfrac\pi2[/tex], so it's only a valid inverse of [tex]\tan x[/tex] if we restrict the domain of [tex]\tan x[/tex] to [tex]-\dfrac\pi2<x<\dfrac\pi2[/tex].
In other words, [tex]x[/tex] is some angle whose terminus lies in either the first or fourth quadrant (I'm assuming you're at least a little familiar with the typical unit circle diagrams used in trig). The only value of [tex]x[/tex] in this domain that satisfies the equation above is [tex]x=-\dfrac\pi3[/tex].
On the other hand, [tex]\mathrm{Arctan}(-\sqrt 3)[/tex] captures all possible values of [tex]x[/tex] that satisfy
[tex]\tan x=-\sqrt3[/tex]
First, we know that [tex]\tan(x+\pi)=\tan x[/tex] (in other words, [tex]\tan[/tex] has a period of [tex]\pi[/tex]), so in fact [tex]\tan(x+2\pi)=\tan(x-2\pi)=\cdots=\tan x[/tex] as well. In general, then, we know that [tex]x+n\pi[/tex] will be a valid solution to the equation above, where [tex]n[/tex] is any integer.
Earlier we found that [tex]\arctan(-\sqrt3)=-\dfrac\pi3[/tex], so in this case, [tex]\mathrm{Arctan}(-\sqrt3)=-\dfrac\pi3+n\pi[/tex] for [tex]n\in\mathbb Z[/tex].
