3.314. $\frac{\cos 2 \alpha-\sin 4 \alpha-\cos 6 \alpha}{\cos 2 \alpha+\sin 4 \alpha-\cos 6 \alpha}$.
## Solution.

$$
\begin{aligned}
& \frac{\cos 2 \alpha - \sin 4 \alpha - \cos 6 \alpha}{\cos 2 \alpha + \sin 4 \alpha - \cos 6 \alpha} = \frac{(\cos 2 \alpha - \cos 6 \alpha) - \sin 4 \alpha}{(\cos 2 \alpha - \cos 6 \alpha) + \sin 4 \alpha} = \\
& = \left[\cos x - \cos y = -2 \sin \frac{x+y}{2} \sin \frac{x-y}{2}\right] = \frac{-2 \sin 4 \alpha \sin (-2 \alpha) - \sin 4 \alpha}{-2 \sin 4 \alpha \sin (-2 \alpha) + \sin 4 \alpha} = \\
& = \frac{2 \sin 4 \alpha \sin 2 \alpha - \sin 4 \alpha}{2 \sin 4 \alpha \sin 2 \alpha + \sin 4 \alpha} = \frac{2 \sin 4 \alpha \left(\sin 2 \alpha - \frac{1}{2}\right)}{2 \sin 4 \alpha \left(\sin 2 \alpha + \frac{1}{2}\right)} = \frac{\sin 2 \alpha - \sin 30^{\circ}}{\sin 2 \alpha + \sin 30^{\circ}} = \\
& = \left[\sin x - \sin y = 2 \cos \frac{x+y}{2} \sin \frac{x-y}{2}\right. \\
& \left.\sin x + \sin y = 2 \sin \frac{x+y}{2} \cos \frac{x-y}{2}\right] = \\
& = \frac{2 \cos \left(\alpha + 15^{\circ}\right) \sin \left(\alpha - 15^{\circ}\right)}{2 \sin \left(\alpha + 15^{\circ}\right) \cos \left(\alpha - 15^{\circ}\right)} = \tan\left(\alpha - 15^{\circ}\right) \cot\left(\alpha + 15^{\circ}\right)
\end{aligned}
$$

Answer: $\tan\left(\alpha - 15^{\circ}\right) \cot\left(\alpha + 15^{\circ}\right)$