Prove that - cos 2-cos 2- -sin 2 - - -sin 2 - - -cos 2 - -

LHS =cos 2θ #160; cos 2ϕ #160; +[ sin( θ ϕ +θ +ϕ #160; ) #160; sin( θ ϕ θ +ϕ #160; ) #160; #160; ] =cos 2θ #160; cos 2ϕ #160; ...

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Algebraic Identities

Prove that : ...

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Prove that : $cos 2 θ cos 2 ϕ + {sin}^{2} (θ - ϕ) - {sin}^{2} (θ + ϕ) = cos 2 (θ + ϕ)$

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cos 2 θ cos 2 ϕ + {sin}^{2} (θ - ϕ) - {sin}^{2} (θ + ϕ) = cos (2 θ + 2 ϕ)

cos 2 θ cos 2 ϕ + {sin}^{2} (θ - ϕ) - {sin}^{2} (θ + ϕ) = cos (2 θ + 2 ϕ)

{tan}^{2} ϕ - {sin}^{2} ϕ = {tan}^{2} ϕ . {sin}^{2} ϕ

{tan}^{2} θ = 2 {tan}^{2} ϕ + 1

cos 2 θ + 2 {sin}^{2} ϕ =

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