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Prove that sqrt((1 – sin θ)/(1 + sin θ)) = sec θ – tan θ.

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Question

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

Theorem
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Solution 1

L.H.S. = `sqrt(((1 - sin θ)(1 - sin θ))/((1 + sin θ)(1 - sin θ)))`

= `sqrt((1 + sin^2θ - 2sinθ)/(1 - sin^2θ)`

= `sqrt((1 + sin^2θ - 2sinθ)/(cos^2θ)`

= `sqrt( 1/cos^2θ + sin^2θ/cos^2θ - (2sin θ)/cos θ xx 1/cosθ`

= `sqrt( sec^2θ + tan^2 θ - 2 tan θ . sec θ)`

= `sqrt((sec θ - tan θ)^2)`

= sec θ – tan θ

= R.H.S.

Hence proved.

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Solution 2

L.H.S. = `sqrt((1 - sin θ)/(1 + sin θ))`

= `sqrt(((1 - sin θ)(1 - sin θ))/((1 + sin θ)(1 - sin θ))`

= `sqrt(((1 - sin θ)^2)/(1 - sin^2θ)`

= `sqrt(((1 - sin θ)^2)/(cos^2θ)`

= `(1 - sin θ)/(cos θ)`

= `1/(cos θ) - (sin θ)/(cos θ)`

= sec θ – tan θ

= R.H.S.

Hence Proved.

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2019-2020 (March) Basic - Delhi set 1

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