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Maharashtra State BoardSSC (English Medium) 10th Standard

Prove that (1 + sin θ)/(1 - sin θ) = (sec θ + tan θ)^2.

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Question

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

Theorem
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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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Chapter 6: Trigonometry - Exercise

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