4.61 $\frac{\cos \left(4 \alpha-\frac{\pi}{2}\right) \sin \left(\frac{5 \pi}{2}+2 \alpha\right)}{(1+\cos 2 \alpha)(1+\cos 4 \alpha)}$.
4.61 Using reduction formulas first, and then (4.18) and (4.17), we find

$\frac{\cos \left(4 \alpha-\frac{\pi}{2}\right) \sin \left(\frac{5 \pi}{2}+2 \alpha\right)}{(1+\cos 2 \alpha)(1+\cos 4 \alpha)}=\frac{\sin 4 \alpha \cos 2 \alpha}{(1+\cos 2 \alpha)(1+\cos 4 \alpha)}=$

$=\frac{\sin 2 \alpha}{1+\cos 2 \alpha} \cdot \frac{2 \cos ^{2} 2 \alpha}{1+\cos 4 \alpha}=\tan \alpha$.