3.243. $(\cos 8 \alpha \tan 4 \alpha-\sin 8 \alpha)(\cos 8 \alpha \cot 4 \alpha+\sin 8 \alpha)$.
## Solution.

$$
\begin{aligned}
& (\cos 8 \alpha \tan 4 \alpha - \sin 8 \alpha)(\cos 8 \alpha \cot 4 \alpha + \sin 8 \alpha)= \\
& =\left(\frac{\cos 8 \alpha \sin 4 \alpha}{\cos 4 \alpha} - \sin 8 \alpha\right)\left(\frac{\cos 8 \alpha \cos 4 \alpha}{\sin 4 \alpha} + \sin 8 \alpha\right)= \\
& =\frac{\sin 4 \alpha \cos 8 \alpha - \cos 4 \alpha \sin 8 \alpha}{\cos 4 \alpha} \cdot \frac{\cos 8 \alpha \cos 4 \alpha + \sin 8 \alpha \sin 4 \alpha}{\sin 4 \alpha}= \\
& =\frac{\sin (-4 \alpha)}{\cos 4 \alpha} \cdot \frac{\cos 4 \alpha}{\sin 4 \alpha}=\frac{-\sin 4 \alpha \cos 4 \alpha}{\sin 4 \alpha \cos 4 \alpha}=-1 .
\end{aligned}
$$

Answer: -1 .