Answer
See proof below.
Work Step by Step
Apply the sum and difference formulas:
$\cos(A+B)=\cos(A)\cos(B) -\sin(A)\sin(B)$
and
$\cos(A-B)=\cos(A)\cos(B) +\sin(A) \ \sin(B)$
In order to prove the given identity, we simplify the left hand side $\text{LHS}$ as follows:
$\text{LHS}=\cos(a-b) \ cos(a+b) \\=[\cos(a)\cos(b) +\sin(a) \ \sin(b)] \ [\cos(a)\cos(b) -\sin(a)\sin(b)] \\(\cos a \ cos b )^2 -(\sin a \cdot \sin b)^2\\=\cos^2 a \cos^2 b -\sin^2 a \sin^2 b\\=\cos^2 a(1-\sin^2 b) -(1-\cos^2 a) \sin^2 b \\=\cos^2 a-\sin^2 b \\ =\text{ RHS}$
