$4.27 \frac{1}{\cos ^{4} \alpha}-\frac{1}{\sin ^{4} \alpha}$.
### 4.27 We have

$$
\begin{aligned}
& \frac{1}{\cos ^{4} \alpha}-\frac{1}{\sin ^{4} \alpha}=\frac{\sin ^{4} \alpha-\cos ^{4} \alpha}{\sin ^{4} \alpha \cos ^{4} \alpha}= \\
& =\frac{\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)\left(\sin ^{2} \alpha-\cos ^{2} \alpha\right)}{\frac{1}{16} \sin ^{4} 2 \alpha}=-\frac{16 \cos 2 \alpha}{\sin ^{4} 2 \alpha}= \\
& =-\frac{16 \operatorname{ctg} 2 \alpha}{\sin ^{3} 2 \alpha}
\end{aligned}
$$

Here, formulas (4.13), (4.14), (4.1), and (4.3) were used.