Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$

For a slightly different solution,

$2\cos^2 A-2\sin^2 B$

$=2\cos^2 A-1+1-2\sin^2 B$

$=\cos 2A+\cos 2B$

$=2\cos (A+B) \cos (A-B)$

and halve each side.

Maybe you didn't see you can use the formula $(x+y)(x-y)=x^2-y^2$. Other than that, I don't see any other simpler method.

You can use the addition theorem which states that $$\cos(\alpha+\beta)=\cos(\alpha)\cdot \cos(\beta) -\sin(\alpha)\sin(\beta)$$

$$\cos(\alpha + \beta) \cdot \cos(\alpha -\beta)= ( \cos(\alpha)\cos(\beta) -\sin(\alpha)\sin(\beta))\cdot (\cos(\alpha) \cdot \cos(-\beta)-\sin(\alpha)\cdot \sin(-\beta))$$ As $\cos(x)=\cos(-x)$ and $\sin(x)=-\sin(-x)$ this is equal to $$(\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta))\cdot (\cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta))$$ With the third binom you get $$\cos^2(\alpha) \cos^2(\beta) -\sin^2(\alpha)\sin^2(\beta)$$ As $\cos^2(\beta)+\sin^2(\beta)=1$ we have $$\cos^2(\alpha)(1-\sin^2(\beta))-(1-\cos^2(\alpha))\sin^2(\beta)$$ Multplying it out gives you $$\cos^2(\alpha) -\sin^2(\beta)\cos^2(\alpha)-\sin^2(\beta)+\cos^2(\alpha)\sin^2 (\beta)$$ And this is $$\cos^2(\alpha)-\sin^2(\beta)$$

Another way: $$2\cos(\alpha)\cdot \cos(\beta)=\cos(\alpha+\beta) + \cos(\alpha-\beta)$$ Using this in your formula gives us $$\cos(A+B)\cdot \cos(A-B)=\frac{1}{2}\left( \cos(2A) + \cos(-2B)\right)=\frac{1}{2} \left(\cos(2 A)+ \cos(2 B)\right)$$ As $\cos(2 A)=1-2\sin^2(A)$ and $\cos(2 B) = 1-2 \sin^2 (B)$ this is equal to $$1-\sin^2(A) -\sin^2 (B)=\cos^2(A)-\sin^2(B)$$