= Solution
The <angular momentum commutation relations> imply
$$
[S_z,S_\pm]=\pm\hbar S_\pm,
\qquad [\mathbf S^2,S_\pm]=0.
$$
Thus $S_\pm|s,\sigma\rangle$ has the same total-spin <eigenvalue> and magnetic quantum number $\sigma\pm1$. Its squared norm follows from
$$
S_\mp S_\pm=\mathbf S^2-S_z^2\mp\hbar S_z:
$$
$$
\lVert S_\pm|s,\sigma\rangle\rVert^2
=\hbar^2\left[s(s+1)-\sigma(\sigma\pm1)\right].
$$
Choosing the conventional positive phases gives the <spin ladder operator> formula
$$
\boxed{S_\pm|s,\sigma\rangle
=\hbar\sqrt{s(s+1)-\sigma(\sigma\pm1)}
|s,\sigma\pm1\rangle}.
$$
For $s=1$, in the ordered basis $|1,1\rangle,|1,0\rangle,|1,-1\rangle$,
$$
S_+=\sqrt2\hbar
\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix},
\qquad S_-=S_+^\dagger.
$$
Since $S_x=(S_++S_-)/2$,
$$
\boxed{S_x=\frac{\hbar}{\sqrt2}
\begin{pmatrix}0&1&0\\1&0&1\\0&1&0\end{pmatrix}}.
$$
The incident state is the <spin coherent state> obtained by rotating $|1,1\rangle$ through $\theta$ about the $y$ axis. Its amplitudes in the $S_z$ basis are
$$
\langle1,1|\mathbf n,+\rangle=\cos^2\frac\theta2,
\qquad
\langle1,0|\mathbf n,+\rangle=\frac{\sin\theta}{\sqrt2},
\qquad
\langle1,-1|\mathbf n,+\rangle=\sin^2\frac\theta2,
$$
up to an irrelevant phase convention. The <Born rule> therefore gives
$$
\boxed{\frac{N_+}{N_-}
=\frac{\cos^4(\theta/2)}{\sin^4(\theta/2)}
=\cot^4\frac\theta2}.
$$
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