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Question

Prove that:
sec(π4+θ)sec(π4θ)=2sec2θ

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Solution

LHS=sec(π4+θ)sec(π4θ)=1cos(π4θ)×1cos(π4+θ)22cos(π4θ)cos(π4+θ)=2cos(2π4)+cos(2θ)=2cos(π2)+cos(2θ)=20+cos(2θ)=2cos(2θ)=2sec2θ=RHS
Hence L.H.S=R.H.S

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