= Solution
For a <Lorentz boost> of speed $v$ along $x$, the <velocity-addition formula> formulas are
$$
u'_x=\frac{u_x-v}{1-u_xv/c^2},\qquad
u'_y=\frac{u_y}{\gamma_v(1-u_xv/c^2)}.
$$
Substitute $u_x=u\cos\theta$, $u_y=u\sin\theta$ and take the ratio. The common time-transformation denominator cancels, yielding
$$
\boxed{\tan\theta'=\frac{u\sin\theta}{\gamma_v(u\cos\theta-v)}.}
$$
To specify the direction rather than just its tangent, use $\theta'=\operatorname{atan2}(u\sin\theta/\gamma_v,u\cos\theta-v)$. This keeps the correct quadrant when the boost reverses the sign of the longitudinal <velocity>. For physical subluminal speeds the canceled denominator is positive. At $u_x=v$ and nonzero $u_y$, the transformed direction is perpendicular to the boost; if the particle is exactly comoving, its transformed velocity vanishes and the angle is undefined.
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