Angle of rotation of an ellipsoid in a linear shear flow field

I am modeling the motion of an ellipsoid in a linear shear flow field. The ellipsoid is rotating about its shortest semi-principal axis which I have designated the $z$-axis in the body-fixed frame, and so we can consider its motion in the $x$-$y$ plane only. I have derived an expression for $\phi(t)$, the angle of rotation of the ellipsoid in this plane, as a function of time. $\phi$ has a range of $\pi$ and is discontinuous. This makes sense since

$$\phi(t) = \arctan\left( \sqrt{\frac{1+r}{1-r}}\tan \left(\frac{\gamma}{2} \sqrt{1-r^2} t\right)\right).$$

where $0 < r < 1$ is a rational expression involving principal axes and $\gamma$ is the constant fluid shear rate along the $y$ axis.

Physically, the ellipsoid should rotate through the entire $2\pi$ (it is tumbling in the flow). The angle $\phi$ is in fact the Euler angle lying in this plane, which should have a range mod $2\pi$.

Is it possible to identify the angle $-\pi/2$ with $\pi/2$ since the ellipsoid is symmetric? Which is to say, is it possible that my function can actually represent an ellipsoid rotating through a full $2 \pi$, or do I have to assume that this function describes discontinuous motion through $\pi$ rads (and is thus wrong)?

For brevity I have excluded a discussion of the derivation of $\phi$, which I have obtained by following two published results. If it would be useful I can include a derivation and the references.

The direct answer to you question is: No, you may not in general identify $\phi=-\pi/2$ with $\phi=\pi/2$. The points to not correspond to the same physical conditions. The exception is if you are in fact treating a two-dimensional particle (an ellipse) in a two-dimensional world. But as the question stands that does not seem to be the case.
a) The equation in the question is valid for ellipsoids of revolution, spheroids. The angle $\phi$ is the azimuthal angle in a spherical coordinate system describing the axis of revolution of the spheroid.
b) The original question arises because of the inverse tangent in the solution. The original derivation of the equation, due to Jeffery (1922) DOI link (pdf), instead states (Eq. 48): $$\tan\phi=\frac{a}{b}\tan\frac{\kappa abt}{a^2+b^2}.$$ This leaves no ambiguity. The function $\phi(t)$ is a solution of the equation of motion (Eq. 47): $$(a^2+b^2)\dot\phi = \kappa(a^2\cos^2\phi + b^2\sin^2\phi).$$ If this equation is integrated numerically one finds that $\phi(t)$ grows monotonically (for physical values of the parameters).
c) Depending on what application you have in mind, you may consider working directly with the cartesian components of the unit vector $\mathbf n(t)$, instead of the spherical parametrization. The corresponding equation of motion is $$\dot{\mathbf n}=\mathbb O\mathbf n + \Lambda(\mathbb S\mathbf n-\mathbf n \mathbf n^T\mathbb S \mathbf n)$$ where the constant flow-gradient matrices $\mathbb O$ and $\mathbb S$ represent the flow, and the constant $\Lambda$ represents the shape of the particle (related to $r$ in question). This equation is, perhaps surprisingly, exactly solvable by a simple matrix exponential: $$\mathbf n(t)=\frac{\exp[(\mathbb O + \Lambda \mathbb S)t]\mathbb n(0)}{|\exp[(\mathbb O + \Lambda \mathbb S)t]\mathbb n(0)|}$$