<s>[INST] 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.
 [/INST] \begin{align}\frac{\sin\frac{m+1}{2}\theta\cos^2\frac{m-1}{2}\theta}{\sin\frac{m-1}{2}\theta}+\frac{\sin\frac{m-1}{2}\theta\cos^2\frac{m+1}{2}\theta}{\sin\frac{m+1}{2}\theta}&=\frac{\left( \sin \frac{m+1}2\theta \cos \frac{m-1}2\theta\right)^2+\left( \sin \frac{m-1}2\theta\cos\frac{m+1}2\theta\right)^2}{\sin \frac{m-1}2\theta \sin \frac{m+1}2\theta}\\
&=\frac{\left( \sin \frac{m+1}2\theta \cos \frac{m-1}2\theta\right)^2+\left( \sin \frac{m-1}2\theta\cos\frac{m+1}2\theta\right)^2}{\sin \frac{m-1}2\theta \sin \frac{m+1}2\theta}\\
&=\frac{\left( \sin m\theta +\sin \theta\right)^2+\left( \sin m\theta-\sin\theta\right)^2}{4\sin \frac{m-1}2\theta \sin \frac{m+1}2\theta}\\
&=\frac{\sin^2 m\theta + \sin^2 \theta}{{2\sin \frac{m-1}2\theta \sin \frac{m+1}2\theta}} \\
&=\frac{\sin^2 m\theta + \sin^2 \theta}{\cos \theta- \cos m\theta}
\end{align}

From this desmos link, the vertical asymptotes is a function of $m$. Hence $m$ is a term in the denominator.
The vertical asymptotes (which due to dision by $0$ in the denominator) appear where $\sin \left( \frac{m-1}2 \theta\right)=0$ or $\sin \left( \frac{m+1}2 \theta\right)=0$, that is when 
$$\frac{m \pm 1}2 \theta = k\pi, k \in \mathbb{Z}$$
$$\theta = \frac{2k\pi}{m \pm 1} , k \in \mathbb{Z}$$
which is a function of $m$. The denominator is dependent on $m$.
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