\begin{align} \sin^2\alpha -\sin^2\beta &= (\sin\alpha+\sin\beta)(\sin\alpha-\sin\beta)\\ &= \Big(2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}\Big)\Big(2\cos\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}\Big)\\ &=\Big(2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha+\beta}{2}\Big)\Big(2\sin\frac{\alpha-\beta}{2}\cos\frac{\alpha-\beta}{2}\Big)\\ &=sin(\alpha+\beta)\sin(\alpha-\beta) \end{align} \begin{align} \cos^2\alpha - \cos^2\beta &= (\cos\alpha + \cos\beta)(\cos\alpha -\cos\beta)\\ &=\Big(2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}\Big)\Big(-2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}\Big)\\ &=-\Big(2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha+\beta}{2}\Big)\Big(2\sin\frac{\alpha-\beta}{2}\cos\frac{\alpha-\beta}{2}\Big)\\ &=-\sin(\alpha+\beta)\sin(\alpha-\beta) \end{align} Note, it is easy to show when adding the two to-be-proved equations, the left sides equals $0$. Therefore after having proved the $1^{st}$ equation, we can simply claim the right side of the $2^{nd}$ equation must be the opposite to the right side of the $1^{st}$ equation.