$4.10 \cos ^{2}\left(45^{\circ}-\alpha\right)-\cos ^{2}\left(60^{\circ}+\alpha\right)-\cos 75^{\circ} \sin \left(75^{\circ}-2 \alpha\right)=\sin 2 \alpha$.
4.10 Let's use formulas (4.17) and (4.27):

$$
\begin{aligned}
& \cos ^{2}\left(45^{\circ}-\alpha\right)-\cos ^{2}\left(60^{\circ}+\alpha\right)-\cos 75^{\circ} \sin \left(75^{\circ}-2 \alpha\right)= \\
& =\frac{1+\cos \left(90^{\circ}-2 \alpha\right)}{2}-\frac{1+\cos \left(120^{\circ}-2 \alpha\right)}{2}- \\
& -\frac{\sin \left(150^{\circ}-2 \alpha\right)+\sin (-2 \alpha)}{2}= \\
& =\frac{\sin 2 \alpha+\sin 2 \alpha-\cos \left(120^{\circ}+2 \alpha\right)-\sin \left(150^{\circ}-2 \alpha\right)}{2}=\sin 2 \alpha
\end{aligned}
$$

Indeed, the third and fourth terms in the numerator cancel each other out, since

$$
\begin{aligned}
& \sin \left(150^{\circ}-2 \alpha\right)=\sin \left(90^{\circ}+\left(60^{\circ}-2 \alpha\right)\right)=\cos \left(60^{\circ}-2 \alpha\right), \text{ and} \\
& \cos \left(120^{\circ}+\alpha\right)=\cos \left(180^{\circ}-\left(60^{\circ}-2 \alpha\right)\right)=-\cos \left(60^{\circ}-2 \alpha\right)
\end{aligned}
$$