$4.17 \frac{(1-\cos 2 \alpha) \cos \left(45^{\circ}+\alpha\right)}{2 \sin ^{2} 2 \alpha-\sin 4 \alpha}$.
4.17 Using formulas (4.16) and (4.13), we have

$$
\begin{aligned}
& A=\frac{2 \sin ^{2} \alpha \cos \left(45^{\circ}+2 \alpha\right)}{2 \sin 2 \alpha(\sin 2 \alpha-\cos 2 \alpha)}= \\
& =\frac{\sin ^{2} \alpha \cos \left(45^{\circ}+2 \alpha\right)}{2 \sin \alpha \cos \alpha(\sin 2 \alpha-\cos 2 \alpha)}=\frac{\operatorname{tg} \alpha \cos \left(45^{\circ}+2 \alpha\right)}{2(\sin 2 \alpha-\cos 2 \alpha)}
\end{aligned}
$$

We transform the difference $\sin 2 \alpha-\cos 2 \alpha$ as follows:

$$
\begin{aligned}
& \sin 2 \alpha-\cos 2 \alpha=\sqrt{2}\left(\frac{1}{\sqrt{2}} \sin 2 \alpha-\frac{1}{\sqrt{2}} \cos 2 \alpha\right)= \\
& =\sqrt{2}\left(\sin 45^{\circ} \sin 2 \alpha-\cos 45^{\circ} \cos 2 \alpha\right)=-\sqrt{2} \cos \left(45^{\circ}+2 \alpha\right)
\end{aligned}
$$

As a result, we obtain

$$
A=-\frac{\operatorname{tg} \alpha \cos \left(45^{\circ}+2 \alpha\right)}{2 \sqrt{2} \cos \left(45^{\circ}+2 \alpha\right)}=-\frac{\sqrt{2}}{4} \operatorname{tg} \alpha
$$