$\begin{array}{l}\text { Example 5. Find } S_{n}=\frac{1}{\sin \theta \sin 3 \theta}+\frac{1}{\sin \theta \sin 5 \theta} \\ +\cdots+\frac{1}{\sin (2 n-1) \theta \sin (2 n+1) \theta} .\end{array}$
$$
\begin{array}{l}
\text { Solve } u_{1}(\theta) \sin 2 \theta=\frac{\sin 2 \theta}{\sin \theta \sin 3 \theta} \\
=\frac{\sin (3 \theta-\theta)}{\sin \theta \sin 3 \theta} \\
=\frac{\sin 3 \theta \cos \theta-\sin \theta \cos 3 \theta}{\sin \theta \sin 3 \theta} \\
=\cot \theta-\cot 3 \theta \text {. } \\
\end{array}
$$

Similarly, for $u_{k}(\theta) \sin 2 \theta$
$$
\begin{array}{l}
=\frac{\sin [(2 k+1)-(2 k-1)] \theta}{\sin (2 k-1) \theta \sin (2 k+1) \theta} \\
=\cot(2 k-1) \theta-\cot(2 k+1) \theta .
\end{array}
$$

Therefore, $\sin 2 \theta \sum_{k=1}^{n} u_{k}(\theta)=\cot \theta-\cot(2 n+1) \theta$.
$$
S_{n}^{(\theta)}=\frac{\cot \theta-\cot(2 n+1) \theta}{\sin 2 \theta} .
$$