To Prove sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x

RHS = cos x

LHS = sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x

LHS = � [2 sin (n + 1)x sin (n + 2)x + 2 cos (n + 1)x cos (n + 2)x]

[Since, 2 sin A sin B = cos (A – B) – cos (A + B)

2 cos A cos B = cos (A + B) + cos (A – B)]

LHS = � [cos {(n + 1)x – (n + 2)x} - cos {(n + 1)x + (n + 2)x} + cos {(n + 1)x + (n + 2)x} + cos {(n + 1)x – (n + 2)x}]

LHS = � [cos (nx + x – nx – 2x) – cos (nx + x + nx + 2x) + cos (nx + x + nx + 2x) + cos (nx + x – nx – 2x)]

LHS = � [cos (-x) – cos (2nx + 3x) + cos (2nx + 3x) + cos (-x)]

LHS = � × 2 cos (-x)

LHS = cos (-x) = cos x

∴ LHS = RHS

Hence, proved