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Prove that (Sec θ - 1)/(Sec θ + 1) = ((Sin θ)/(1 + Cos θ ))^2 - Mathematics

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Question

Prove that `(sec θ - 1)/(sec θ + 1) = ((sin θ)/(1 + cos θ ))^2`

Sum
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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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