$4.6 \frac{\cos \left(\frac{5 \pi}{2}-6 \alpha\right)+\sin (\pi+4 \alpha)+\sin (3 \pi-\alpha)}{\sin \left(\frac{5 \pi}{2}+6 \alpha\right)+\cos (4 \alpha-2 \pi)+\cos (\alpha+2 \pi)}=\tan \alpha$.
4.6 First, we will use the reduction formulas and write the left side of the identity as

$\frac{\sin 6 \alpha-\sin 4 \alpha+\sin \alpha}{\cos 6 \alpha+\cos 4 \alpha+\cos \alpha}$.

Now we will use formulas (4.20) and (4.21):

$\frac{2 \sin \alpha \cos 5 \alpha+\sin \alpha}{2 \cos 5 \alpha \cos \alpha+\cos \alpha}=\frac{\sin \alpha(2 \cos 5 \alpha+1)}{\cos \alpha(2 \cos 5 \alpha+1)}=\operatorname{tg} \alpha$.