Relativistic Transverse Doppler Effect In Minkowski spacetime, two observers, $A$ and $B$, are moving at uniform speeds $u$ and $v$, respectively, along different trajectories, each parallel to the y-axis of some inertial frame $S$. Observer $A$ emits a photon with frequency $\nu_{A}$ that travels in the x-direction in $S$ and is received by observer $B$ with frequency $\nu_B$. Show that the Doppler shift $\frac{\nu_B}{\nu_A}$ in the photon frequency is independent of whether $A$ and $B$ are travelling in the same direction or opposite directions. 
Relevant equations: $$\frac{\lambda}{\lambda'} = \frac{\nu_B}{\nu_A} = \gamma(1-\beta\cos\theta)$$
Aberration formula: 
$$\cos\theta' = \frac{\cos\theta - \beta}{1-\beta\cos\theta} = -\beta$$
(for transverse case)
The answer is apparently that  the Doppler shift is independent of the relative direction of motion. I have tried to transform to the frame $S'$ where $B$ is stationary, finding the velocity of $A$ using the addition of velocities formula - to then get gamma. I have used the abberation formula to insert $\cos\theta'$ into the Doppler shift formula above to get $\frac{\nu_B}{\nu_A} = \gamma(1+\beta^2)$. Plugging in the velocity of the emitter $A$ in frame $S'$ doesn't seem to get the required result. 
 A: Before answering your question, please be informed that I use the following correct form of your first equation:
$$\nu_B=\nu_A \frac{1+\frac{w}{c}\cos\theta_A}{\sqrt{1-\frac{w^2}{c^2}}}\space ,$$
where $w=\frac{u+v}{1+{uv}/{c^2}}$ is the relativistic velocity of $A$ as measured by $B$. (Assume that $A$ and $B$ recede from each other along $y_A$ and $y_B$.) However, remember that using the aberration formula, we should find the original angle of emission in $A$'s frame of reference, while the emission angle is calculated as $\theta_S=\pi/2$ in $\boldsymbol S$. Therefore, we must consider $A$'s velocity of $u$ WRT $S$, rather than $w$:
$$\cos \theta_A=\frac{\cos \theta_S-\frac{u}{c}}{1-\frac{u}{c}\cos \theta_S}=-\frac{u}{c} \space.$$
Subtituting the second equation in the first one and using $w=\frac{u+v}{1+{uv}/{c^2}}$ implies:
$$\nu_B=\nu_A \frac{1-\frac{wu}{c^2}}{\sqrt{1-\frac{w^2}{c^2}}}=\nu_A × \frac{c^2-\frac{u(u+v)}{1+uv/c^2}}{c\sqrt{c^2-\frac{u^2+v^2+2uv}{(1+uv/c^2)^2}}}=\nu_A \frac{c^2-u^2}{c\sqrt{c^2+\frac{u^2v^2}{c^2}-u^2-v^2}}=\nu_A \sqrt{\frac{c^2-u^2}{c^2-v^2}}\space .$$
As you see, if you change $u$ into $-u$, or $v$ into $-v$, the result remains the same.
