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Question
Prove that `(sec θ - 1)/(sec θ + 1) = ((sin θ)/(1 + cos θ ))^2`
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Solution
LHS = `(sec θ - 1)/(sec θ + 1)`
= `(1/cos θ - 1)/(1/cos θ + 1)`
= `(1 - cos θ)/(1 + cos θ)`
= `(1 - cos θ xx ( 1 + cos θ))/(1 + cos θ xx (1 + cos θ))`
= `(1 - cos^2 θ)/(1 + cos θ)^2`
= `(sin^2 θ)/(1 + cos θ)^2`
= `((sin θ)/(1 + cos θ ))^2`
= RHS
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