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:
$\text{LHS }=\cos{(\alpha+\beta)}+\cos{(\alpha-\beta)}\\=[\cos{(\alpha)} \ \cos{(\beta)}-\sin{(\alpha)}\sin{(\beta)}]+[(\cos{(\alpha)} \ \cos{(\beta)}+\sin{(\alpha)} \ \sin{(\beta))}]\\=2 \ \cos{(\alpha)} \ \cos{(\beta)} \\=\text{RHS }$
