cos 2θ cos 2<span

Given that, cos 2θ cos 2ϕ + sin²(θ–ϕ) – sin²(θ + ϕ)

= cos 2θ cos 2ϕ + sin(θ–ϕ+ θ + ϕ)sin(θ–ϕ- θ – ϕ)

[∵sin²A – sin²B = sin (A + B) sin (A – B)]

= cos 2θ cos 2ϕ + sin 2θ. sin(- 2ϕ)

= cos 2θ cos 2ϕ - sin 2θ. Sin 2ϕ [∵ sin(-θ) = -sin θ]

= cos 2(θ + ϕ) [∵ cos x cos y – sin x sin y = cos(x + y)]