Answer :

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)]

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