3.294. $\frac{\sin (2 \alpha-\beta)}{\cos 4 \alpha}+\frac{\sin \beta}{\cos 2 \alpha}$.
## Solution.

$$
\begin{aligned}
& \frac{\sin (2 \alpha-\beta)}{\cos 4 \alpha}+\frac{\sin \beta}{\cos 2 \alpha}=\frac{\sin (2 \alpha-\beta) \cos 2 \alpha+\sin \beta \cos 4 \alpha}{\cos 4 \alpha \cos 2 \alpha}= \\
& =\left[\sin x \cos y=\frac{1}{2}(\sin (x-y)+\sin (x+y))\right]= \\
& =\frac{\frac{1}{2}(\sin (-\beta)+\sin (4 \alpha-\beta))+\frac{1}{2}(\sin (\beta-4 \alpha)+\sin (\beta+4 \alpha))}{\cos 4 \alpha \cos 2 \alpha}= \\
& =\frac{-\sin \beta+\sin (4 \alpha-\beta)-\sin (4 \alpha-\beta)+\sin (4 \alpha+\beta)}{2 \cos 4 \alpha \cos 2 \alpha}=\frac{\sin (4 \alpha+\beta)-\sin \beta}{2 \cos 4 \alpha \cos 2 \alpha}= \\
& =\left[\sin x-\sin y=2 \cos \frac{x+y}{2} \sin \frac{x-y}{2}\right]=\frac{2 \cos (2 \alpha+\beta) \sin 2 \alpha}{2 \cos 4 \alpha \cos 2 \alpha}=
\end{aligned}
$$

$=\frac{\cos (2 \alpha+\beta)}{\cos 4 \alpha} \cdot \tan 2 \alpha$.

Answer: $\frac{\cos (2 \alpha+\beta)}{\cos 4 \alpha} \cdot \tan 2 \alpha$.