3.300. $2-\operatorname{tg} 4 \alpha-\operatorname{ctg} 4 \alpha$.

3.300. $2-\tan 4 \alpha-\cot 4 \alpha$.
## Solution.

$$
\begin{aligned}
& 2-\operatorname{tg} 4 \alpha-\operatorname{ctg} 4 \alpha=2-\left(\frac{\sin 4 \alpha}{\cos 4 \alpha}+\frac{\cos 4 \alpha}{\sin 4 \alpha}\right)=2-\frac{\sin ^{2} 4 \alpha+\cos ^{2} 4 \alpha}{\sin 4 \alpha \cos 4 \alpha}= \\
& =2-\frac{1}{\sin 4 \alpha \cos 4 \alpha}=2-\frac{2}{2 \sin 4 \alpha \cos 4 \alpha}=2-\frac{2}{\sin 8 \alpha}=2 \cdot \frac{\sin 8 \alpha-1}{\sin 8 \alpha}= \\
& =2 \cdot \frac{\sin 8 \alpha-\sin \frac{\pi}{2}}{\sin 8 \alpha}=\left[\sin x-\sin y=2 \cos \frac{x+y}{2} \sin \frac{x-y}{2}\right]=
\end{aligned}
$$

$$
=\frac{4 \cos \left(4 \alpha+\frac{\pi}{4}\right) \sin \left(4 \alpha-\frac{\pi}{4}\right)}{\sin 8 \alpha}=\frac{4 \cos \left(\frac{\pi}{2}-\left(\frac{\pi}{4}-4 \alpha\right)\right) \sin \left(-\left(\frac{\pi}{4}-4 \alpha\right)\right)}{\sin 8 \alpha}=
$$

$$
=\frac{4 \sin \left(\frac{\pi}{4}-4 \alpha\right)\left(-\sin \left(\frac{\pi}{4}-4 \alpha\right)\right)}{\sin 8 \alpha}=\frac{-4 \sin ^{2}\left(\frac{\pi}{4}-4 \alpha\right)}{\sin 8 \alpha}
$$

Answer: $\frac{-4 \sin ^{2}\left(\frac{\pi}{4}-4 \alpha\right)}{\sin 8 \alpha}$.