Simplifying $\frac{\sin\alpha\cos^2\beta}{\sin\beta}+\frac{\sin\beta\cos^2\alpha}{\sin\alpha}$ How to simplify the expression and get a closed form?

$$\frac{\sin\alpha\cos^2\beta}{\sin\beta}+\frac{\sin\beta\cos^2\alpha}{\sin\alpha}$$
  where $\alpha := \frac12(m+1)\theta$ and $\beta:=\frac12(m-1)\theta$; and where $m$ is an integer.

I wanna get rid of the denominators or make it a constant without $m$, but don't know how.

Let $2a = (m+1) \theta$ and $2b = (m-1) \theta$ to obtain
$$f = \frac{\sin(a) \, \cos^{2}(b)}{\sin(b)} + \frac{\sin(b) \, \cos^{2}(a)}{\sin(a)}.$$
Now, consider the following:
\begin{align}
f &= \frac{1}{\sin(a) \, \sin(b)} \, \left(\sin^{2}(a) \, \cos^{2}(b) + \sin^{2}(b) \, \cos^{2}(a) \right) 
\end{align}
Using $2 \sin^{2} x = 1 - \cos(2 x)$ and $2 \cos^{2} x = 1 + \cos(2 x)$, then
\begin{align}
f &= \frac{1}{\sin(a) \, \sin(b)} \, \left(\sin^{2}(a) \, \cos^{2}(b) + \sin^{2}(b) \, \cos^{2}(a) \right) \\
&= \frac{1}{4 \sin(a) \sin(b)} \, ( (1-\cos(2a))(1+\cos(2b)) + (1-\cos(2b))(1+\cos(2a)) ) \\
&= \frac{1 - \cos(2a) \cos(2b)}{2 \, \sin(a) \sin(b)} \\
&= \frac{2 - \cos2(a-b) - \cos2(a+b)}{4 \, \sin(a) \sin(b)} \\
&= \frac{\sin^{2}(a-b) + \sin^{2}(a+b)}{2 \, \sin(a) \sin(b)} 
\end{align}
This can be seen in the form
$$f = \frac{\sin^{2}(m \theta) + \sin^{2}\theta}{\cos\theta - \cos(m \theta)}.$$
It becomes evident from $\sin(a) \, \sin(b)$ that there are problems (asymptotes) at 
$$\theta \in \left\{ \frac{2 n \pi}{m \pm 1} \right\} \hspace{5mm} n \geq 0. $$
