3.193.

$$
\frac{2 \cos \left(\frac{\pi}{6}-2 \alpha\right)-\sqrt{3} \sin \left(\frac{5 \pi}{2}-2 \alpha\right)}{\cos \left(\frac{9 \pi}{2}-2 \alpha\right)+2 \cos \left(\frac{\pi}{6}+2 \alpha\right)}=\frac{\tan 2 \alpha}{\sqrt{3}}
$$
## Solution.

$$
\begin{aligned}
& \frac{2 \cos \left(\frac{\pi}{6}-2 \alpha\right)-\sqrt{3} \sin \left(\frac{5 \pi}{2}-2 \alpha\right)}{\cos \left(\frac{9 \pi}{2}-2 \alpha\right)+2 \cos \left(\frac{\pi}{6}+2 \alpha\right)}= \\
= & \frac{2\left(\cos \frac{\pi}{6} \cos 2 \alpha+\sin \frac{\pi}{6} \sin 2 \alpha\right)-\sqrt{3} \cos 2 \alpha}{\sin 2 \alpha+2\left(\cos \frac{\pi}{6} \cos 2 \alpha-\sin \frac{\pi}{6} \sin 2 \alpha\right)}= \\
= & \frac{2\left(\frac{\sqrt{3}}{2} \cos 2 \alpha+\frac{1}{2} \sin 2 \alpha\right)-\sqrt{3} \cos 2 \alpha}{\sin 2 \alpha+2\left(\frac{\sqrt{3}}{2} \cos 2 \alpha-\frac{1}{2} \sin 2 \alpha\right)}=
\end{aligned}
$$

$$
=\frac{\sqrt{3} \cos 2 \alpha+\sin 2 \alpha-\sqrt{3} \cos 2 \alpha}{\sin 2 \alpha+\sqrt{3} \cos 2 \alpha-\sin 2 \alpha}=\frac{\sin 2 \alpha}{\sqrt{3} \cos 2 \alpha}=\frac{\operatorname{tg} 2 \alpha}{\sqrt{3}} .
$$

The identity is proven.