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Question
Prove that `((1 + sin θ - cos θ)/( 1 + sin θ + cos θ))^2 = (1 - cos θ)/(1 + cos θ)`.
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Solution
LHS = `((1 + sin θ - cos θ)/( 1 + sin θ + cos θ))^2`
⇒ `(1 + sin^2 θ + cos^2 θ + 2(sin θ - cos θ - sin θ. cos θ))/(1 + sin^2 θ + cos^2 θ + 2(sin θ + cos θ + sin θ. cos θ)`
= `(1 + 1 + 2 (sin θ - cos θ - sin θ. cos θ))/( 1 + 1 + 2((sin θ + cos θ + sin θ. cos θ)`
= `(2 (1 + sin θ - cos θ - sin θ. cos θ))/(2( 1 + (sin θ + cos θ + sin θ. cos θ))`
= `( 1 + sin θ - cos θ( 1 + sin θ))/(1 + sin θ + cos θ( 1 + sin θ))`
= `((1 + sin θ)(1 - cos θ))/((1 + sin θ)( 1 + cos θ))`
= `(1 - cos θ)/( 1 + cos θ)`
= RHS
Hence proved.
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