= Solution
Let the <inertial frame> $S'$ move at speed $v$ in the positive $x$ direction relative to $S$, with parallel axes and coincident origins at zero time. The inverse <Lorentz transformation> gives
$$
dt=\gamma_v(dt'+v\,dx'/c^2),\quad dx=\gamma_v(dx'+v\,dt'),\quad dy=dy',\quad dz=dz'.
$$
Here $\gamma_v=(1-v^2/c^2)^{-1/2}$ and $\mathbf u'=d\mathbf x'/dt'$ is the particle's <velocity> in $S'$. Dividing spatial differentials by $dt$ gives the <relativistic velocity addition> law
$$
\boxed{u_x=\frac{u_x'+v}{1+vu_x'/c^2},\qquad
u_y=\frac{u_y'}{\gamma_v(1+vu_x'/c^2)},\qquad
u_z=\frac{u_z'}{\gamma_v(1+vu_x'/c^2)}.}
$$
For the collision, set $c=1$ during the algebra. Let the incident projectile have <energy> $E>m_1$ and <momentum> $p=\sqrt{E^2-m_1^2}$. The target is initially at rest. The squared <invariant mass>, boost speed and <Lorentz factor> of the <center-of-momentum frame> are
$$
s=m_1^2+m_2^2+2m_2E,\qquad V=\frac p{E+m_2},\qquad\Gamma=\frac{E+m_2}{\sqrt s}.
$$
In that frame the projectile has $p_*=m_2p/\sqrt s$ and $E_*=(m_1^2+m_2E)/\sqrt s$. An <elastic collision> changes the direction of its <momentum>, through an angle $\varphi$, while preserving these magnitudes. Transforming the outgoing <four-momentum> back to the laboratory gives
$$
\tan\theta=\frac{\sin\varphi}{\Gamma(\cos\varphi+B)},\qquad
B=\frac{VE_*}{p_*}=\frac{m_1^2+m_2E}{m_2(E+m_2)}>1.
$$
The maximum occurs at $\cos\varphi=-1/B$. Since
$$
\Gamma^2(B^2-1)=\frac{m_1^2-m_2^2}{m_2^2},
$$
the <maximum laboratory deflection in relativistic elastic scattering> is
$$
\boxed{\tan\theta_{\max}=\frac{m_2}{\sqrt{m_1^2-m_2^2}},\qquad
\sin\theta\leq\sin\theta_{\max}=\frac{m_2}{m_1}.}
$$
The printed strict inequality needs correction: equality is kinematically attainable. For example, in $c=1$ units take $m_1=2$, $m_2=1$, and incident $E=3$. A projectile scattered through $30^\circ$ with final <energy> $16/7$ has <momentum> $(3\sqrt5/7,\sqrt{15}/7)$. The target then has <energy> $12/7$ and <momentum> $(4\sqrt5/7,-\sqrt{15}/7)$; both outgoing particles satisfy their <mass shells> and conserve total <four-momentum>. This gives $\sin\theta=1/2=m_2/m_1$ exactly. The inequality is strict only away from the maximizing direction.
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