special relativity - Implications of Formula for Stellar Aberration

I am studying relativity and was working on deriving stellar aberration using relativity. My derivation is as follows:

Suppose that the observer is traveling at a velocity $v$ in the $x$-direction. Consider a light beam coming from a star to the ship's crew, which has velocity $c$ and components $u_x=c\cos\theta$ and $u_y=c\sin\theta$. Using the Lorentz transformations, assuming that hte shit is at rest and that the star is moving in the $x$-direction at a velocity of $-v$, we see that $u'_x=\frac{u_x+v}{1+u_xv/c^2}$ and $u_y'=\frac{u_y}{\gamma(1+u_x v/c^2)}$, where $\gamma=\frac{1}{\sqrt{1-(v/c)^2}}$. Then the new angle is given by

$$\tan\theta'=\frac{u_y'}{u'_x}=\frac{u_y}{\gamma(u_x+v)}=\frac{\sin\theta}{\gamma(\cos\theta+(v/c))}.$$

For most angles, this has the desired effect; it makes the angle smaller. However, at the angle $\theta=\arccos(-v/c)$, we encounter a discontinuity, such that the tangent of $\theta'$ is not defined. I am a little hesitant to say that this would thus correspond with $\theta'=\pi/2$, so that $\arccos(-v/c)$ is sent to $\pi/2$. Is this indeed the case?

Additionally, after this (i.e. for angles greater than $\arccos(-v/c)$), $\tan\theta'$ becomes negative. Moreover, for angles $\theta$ near $\arccos(-v/c)$, we can choose $\theta$ close enough to $\arccos(-v/c)$ to be arbitrarily large (in magnitude). But since these angles are naturally modded by $2\pi$, it would seem them that the apparent position of the star could be just about anywhere as observed by the observer. This seems very fishy, and occurs as well with angles just below $\arccos(-v/c)$. How should I interpret this? Indeed, how should I interpret the fact that the angles are negative? Does this means that it is reflected about the $x$-axis?

Note that I am not making (and do not want to make) assumptions about the relative size of $v$; $v$ can be arbitrarily large (while less than $c$).

Thank you for your time.

Answer

Instead of using the tangent, it is better to use the cosine or the sine:

$$\cos\theta' = \frac{u'_x}{c}=\frac{\cos\theta + v/c}{1 + (v/c)\cos\theta},$$ $$\sin\theta' = \frac{u'_y}{c}=\frac{\sin\theta/\gamma}{1 + (v/c)\cos\theta}.$$ Unlike the tangent, this gives you the full information: $\theta'$ is uniquely determined for any value of $\theta$, without discontinuities. And indeed, the angle $\arccos(-v/c)$ is sent to $\pi/2$.

It's also worth noting that the above formulas are valid if one uses the convention that approaching sources have a positive velocity (with $\theta=0$ for a source heading directly towards us). But instead, one often uses the convention that receding sources have a positive velocity (with $\theta=0$ for a source moving directly away from us). In that case, the formulas become

$$\cos\theta' = \frac{\cos\theta - v/c}{1 - (v/c)\cos\theta},$$ $$\sin\theta' = \frac{\sin\theta/\gamma}{1 - (v/c)\cos\theta}.$$ This formula for $\cos\theta'$ corresponds with the formula given in wikipedia.