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Question
Prove that `(1 + sin θ)/(1 - sin θ) = (sec θ + tan θ)^2`.
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Solution
L.H.S. = `(1 + sin θ)/(1 - sin θ)`
= `((1 + sinθ)/(cosθ))/((1 - sinθ)/(cosθ))` ...[Dividing numerator and denominator by cos θ]
= `(1/cosθ + (sinθ)/(cosθ))/(1/cosθ - (sinθ)/(cosθ)`
= `(secθ + tanθ)/(secθ - tanθ)`
= `(secθ + tanθ)/(secθ - tanθ) xx (secθ + tanθ)/(secθ + tanθ)` ...[On rationalising the denominator]
= `(secθ + tanθ)^2/(sec^2θ - tan^2θ)`
= `(secθ + tanθ)^2/1` ...`[(∵ 1 + tan^2θ = sec^2θ),(∴ sec^2θ - tan^2θ = 1)]`
= (sec θ + tan θ)2
= R.H.S.
∴ `(1 + sinθ)/(1 - sinθ) = (sec θ + tan θ)^2`
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