3.040. $\frac{1-\cos 4 \alpha}{\cos ^{-2} 2 \alpha-1}+\frac{1+\cos 4 \alpha}{\sin ^{-2} 2 \alpha-1}=2$.
Solution.

$$
\begin{aligned}
& \frac{1-\cos 4 \alpha}{\cos ^{-2} 2 \alpha-1}+\frac{1+\cos 4 \alpha}{\sin ^{-2} 2 \alpha-1}=\frac{1-\cos 4 \alpha}{\frac{1}{\cos ^{2} 2 \alpha}-1}+\frac{1+\cos 4 \alpha}{\frac{1}{\sin ^{2} 2 \alpha}-1}= \\
& =\frac{(1-\cos 4 \alpha) \cos ^{2} 2 \alpha}{1-\cos ^{2} 2 \alpha}+\frac{(1+\cos 4 \alpha) \sin ^{2} 2 \alpha}{1-\sin ^{2} 2 \alpha}=\frac{(1-\cos 4 \alpha) \cos ^{2} 2 \alpha}{\sin ^{2} 2 \alpha}+ \\
& +\frac{(1+\cos 4 \alpha) \sin ^{2} 2 \alpha}{\cos ^{2} 2 \alpha}=\frac{\left(1-\left(1-\sin ^{2} 2 \alpha\right) \cos ^{2} 2 \alpha\right.}{\sin ^{2} 2 \alpha}+\frac{\left(1+2 \cos ^{2} 2 \alpha-1\right) \sin ^{2} 2 \alpha}{\cos ^{2} 2 \alpha}= \\
& =\frac{2 \sin ^{2} 2 \alpha \cos ^{2} 2 \alpha}{\sin ^{2} 2 \alpha}+\frac{2 \cos ^{2} 2 \alpha \sin ^{2} 2 \alpha}{\cos ^{2} 2 \alpha}=2 \cos ^{2} 2 \alpha+2 \sin ^{2} 2 \alpha= \\
& =2\left(\cos ^{2} 2 \alpha+\sin ^{2} 2 \alpha\right)=2 .
\end{aligned}
$$

The identity is proven.