3.275. $\sqrt{\frac{2 \sin \alpha-\sin 2 \alpha}{2 \sin \alpha+\sin 2 \alpha}}$, if: a) $0<\alpha<\pi$; b) $\pi<\alpha<2 \pi$.
## Solution.

$$
\begin{aligned}
& \sqrt{\frac{2 \sin \alpha - \sin 2 \alpha}{2 \sin \alpha + \sin 2 \alpha}} = \sqrt{\frac{2 \sin \alpha - 2 \sin \alpha \cos \alpha}{2 \sin \alpha + 2 \sin \alpha \cos \alpha}} = \sqrt{\frac{2 \sin \alpha (1 - \cos \alpha)}{2 \sin \alpha (1 + \cos \alpha)}} = \\
& = \sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}} = \left[\frac{1 - \cos x}{1 + \cos x} = \tan^2 \frac{x}{2}, x \neq \pi + 2 \pi n, n \in \mathbb{Z}\right] = \sqrt{\tan^2 \frac{\alpha}{2}} = \\
& = \begin{cases}
\tan \frac{\alpha}{2}, & \text{if } 0 < \alpha < \pi ; \\
-\tan \frac{\alpha}{2}, & \text{if } \pi < \alpha < 2 \pi
\end{cases}
\end{aligned}
$$

Answer: a) $\tan \frac{\alpha}{2} ;$ b) $-\tan \frac{\alpha}{2}$.