3.250. $\frac{(1+\operatorname{tg} 2 \alpha)^{2}-2 \operatorname{tg}^{2} 2 \alpha}{1+\operatorname{tg}^{2} 2 \alpha}-\sin 4 \alpha-1$.

3.250. $\frac{(1+\tan 2 \alpha)^{2}-2 \tan^{2} 2 \alpha}{1+\tan^{2} 2 \alpha}-\sin 4 \alpha-1$.
Solution.

$$
\begin{aligned}
& \frac{(1+\operatorname{tg} 2 \alpha)^{2}-2 \operatorname{tg}^{2} 2 \alpha}{1+\operatorname{tg}^{2} 2 \alpha}-\sin 4 \alpha-1= \\
& =\frac{1+2 \operatorname{tg} 2 \alpha+\operatorname{tg}^{2} 2 \alpha-2 \operatorname{tg}^{2} 2 \alpha}{1+\operatorname{tg}^{2} 2 \alpha}-\sin 4 \alpha-1= \\
& =\frac{1+2 \operatorname{tg} 2 \alpha-\operatorname{tg}^{2} 2 \alpha}{1+\operatorname{tg}^{2} 2 \alpha}-\sin 4 \alpha-1= \\
& =\left[\sin x=\frac{2 \operatorname{tg} \frac{x}{2}}{1+\operatorname{tg}^{2} \frac{x}{2}}, x \neq(2 n+1) \pi, n \in Z\right]= \\
& =\frac{1+2 \operatorname{tg} 2 \alpha-\operatorname{tg}^{2} 2 \alpha}{1+\operatorname{tg}^{2} 2 \alpha}-\frac{2 \operatorname{tg} 2 \alpha}{1+\operatorname{tg}^{2} 2 \alpha}-1=
\end{aligned}
$$

$$
=\frac{1+2 \operatorname{tg} 2 \alpha-\operatorname{tg}^{2} 2 \alpha-2 \operatorname{tg} 2 \alpha-1-\operatorname{tg}^{2} 2 \alpha}{1+\operatorname{tg}^{2} 2 \alpha}=
$$

$$
=\frac{-2 \operatorname{tg}^{2} 2 \alpha}{1+\operatorname{tg}^{2} 2 \alpha}=-2 \cdot \frac{\frac{\sin ^{2} 2 \alpha}{\cos ^{2} 2 \alpha}}{1+\frac{\sin ^{2} 2 \alpha}{\cos ^{2} 2 \alpha}}=-2 \cdot \frac{\frac{\sin ^{2} 2 \alpha}{\cos ^{2} 2 \alpha}}{\frac{\cos ^{2} 2 \alpha+\sin ^{2} 2 \alpha}{\cos ^{2} 2 \alpha}}=-2 \sin ^{2} 2 \alpha .
$$

Answer: $-2 \sin ^{2} 2 \alpha$.