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Question

cos2θcos2ϕ+sin2(θϕ)sin2(θ+ϕ)=cos(2θ+2ϕ)

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Solution

sin2Asin2B=sin(A+B)sin(AB)(1)
LHS
cos(2θ)cos2ϕ+sin2(θϕ)sin2(θ+ϕ)
A=θϕ
B=θ+ϕ
cos(2θ)cos2ϕ+sin(2θ)sin(2ϕ)
cos(2θ2ϕ) =RHS
Hence proved

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