3.282. $\frac{\sin 8 \alpha+\sin 9 \alpha+\sin 10 \alpha+\sin 11 \alpha}{\cos 8 \alpha+\cos 9 \alpha+\cos 10 \alpha+\cos 11 \alpha} \times$

$\times \frac{\cos 8 \alpha-\cos 9 \alpha-\cos 10 \alpha+\cos 11 \alpha}{\sin 8 \alpha-\sin 9 \alpha-\sin 10 \alpha+\sin 11 \alpha}$.
Solution.

$$
\begin{aligned}
& \frac{\sin 8 \alpha + \sin 9 \alpha + \sin 10 \alpha + \sin 11 \alpha}{\cos 8 \alpha + \cos 9 \alpha + \cos 10 \alpha + \cos 11 \alpha} \cdot \frac{\cos 8 \alpha - \cos 9 \alpha - \cos 10 \alpha + \cos 11 \alpha}{\sin 8 \alpha - \sin 9 \alpha - \sin 10 \alpha + \sin 11 \alpha} = \\
& = \frac{(\sin 11 \alpha + \sin 8 \alpha) + (\sin 10 \alpha + \sin 9 \alpha)}{(\cos 11 \alpha + \cos 8 \alpha) + (\cos 10 \alpha + \cos 9 \alpha)} \times \\
& \times \frac{(\cos 11 \alpha + \cos 8 \alpha) - (\cos 10 \alpha + \cos 9 \alpha)}{(\sin 11 \alpha + \sin 8 \alpha) - (\sin 10 \alpha + \sin 9 \alpha)} = \\
& = \left[\cos x + \cos y = 2 \cos \frac{x + y}{2} \cos \frac{x - y}{2}\right. \\
& \left.\sin x + \sin y = 2 \sin \frac{x + y}{2} \cos \frac{x - y}{2}\right] =
\end{aligned}
$$

$$
\begin{aligned}
& = \frac{2 \sin \frac{19 \alpha}{2} \cos \frac{3 \alpha}{2} + 2 \sin \frac{19 \alpha}{2} \cos \frac{\alpha}{2}}{2 \cos \frac{19 \alpha}{2} \cos \frac{3 \alpha}{2} + 2 \cos \frac{19 \alpha}{2} \cos \frac{\alpha}{2}} \times \\
& \times \frac{2 \cos \frac{19 \alpha}{2} \cos \frac{3 \alpha}{2} - 2 \cos \frac{19 \alpha}{2} \cos \frac{\alpha}{2}}{2 \sin \frac{19 \alpha}{2} \cos \frac{3 \alpha}{2} - 2 \sin \frac{19 \alpha}{2} \cos \frac{\alpha}{2}} = \\
& = \frac{2 \sin \frac{19 \alpha}{2} \left( \cos \frac{3 \alpha}{2} + \cos \frac{\alpha}{2} \right)}{2 \cos \frac{19 \alpha}{2} \left( \cos \frac{3 \alpha}{2} + \cos \frac{\alpha}{2} \right)} \cdot \frac{2 \cos \frac{19 \alpha}{2} \left( \cos \frac{3 \alpha}{2} - \cos \frac{\alpha}{2} \right)}{2 \sin \frac{19 \alpha}{2} \left( \cos \frac{3 \alpha}{2} - \cos \frac{\alpha}{2} \right)} = \\
& = \frac{\sin \frac{19 \alpha}{2}}{\cos \frac{19 \alpha}{2}} \cdot \frac{\cos \frac{19 \alpha}{2}}{\sin \frac{19 \alpha}{2}} = 1
\end{aligned}
$$

Answer: 1.