We have,
$$
\int e^{\alpha x}\left(\frac{1-\beta \sin x}{1-\cos x}\right) d x=-e^x \cot \frac{x}{2}+c
$$
On differentiating by both sides, we get
$$
e^{\alpha x}\left(\frac{1-\beta \sin x}{1-\cos x}\right)=-e^x\left[\cot \frac{x}{2}-\frac{1}{2} \operatorname{cosec}^2 \frac{x}{2}\right]
$$

$$
\begin{aligned}
& =-e^x\left(\frac{\cos \frac{x}{2}}{\sin \frac{x}{2}}-\frac{1}{2 \sin ^2 \frac{x}{2}}\right)=-e^x\left[\frac{2 \sin \frac{x}{2} \cos \frac{x}{2}-1}{2 \sin ^2 \frac{x}{2}}\right] \\
& \Rightarrow \quad e^{\alpha x}\left(\frac{1-\beta \sin x}{1-\cos x}\right)=e^x\left(\frac{1-\sin x}{1-\cos x}\right) \\
& \therefore \quad \alpha=1, \beta=1
\end{aligned}
$$
Hence,
$$
\frac{\alpha^2+\beta^2}{2 \alpha \beta}=\frac{1+1}{2}=1
$$