Just to answer your follow-up:

To clarify what I said, here's a diagram:

Basically, we are given the side lengths parallel to the coordinate axes, 1 and tan θ. We can fill in the length of the hypotenuse using Pythagoras' theorem/trigonometric identity. Recall that for any number x+iy on the complex plane that the tangent of its argument will just be y/x. Hence, the tangent of the angle the hypotenuse makes with the real axis (opposite/adjacent !!) will be tan θ/1 as given. This way of thinking eliminates the requirement of using arctan - which really just simplifies things.

To answer the additional question:

You can't always assume that arctan and tan will cancel each other out - recall that arctan(tan x) = x if and only if \(x \in \left(-\frac{\pi}{2}, \ \frac{\pi}{2}\right)\). Here, this assumption is okay because the domain we're given is a subset of the domain for which this identity holds.

Something to also note is that this problem is simplified a lot because of the assumption that θ lies in the first quadrant. This means that every trigonometric ratio is positive and we don't have to worry about minus signs or anything. Basically, it means we don't really have to consider the domain of any of the functions to a rigorous degree.