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प्रश्न
Prove that `sqrt((1 - sin θ)/(1 + sin θ)) = sec θ - tan θ`.
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उत्तर १
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.
उत्तर २
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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Prove that `sec^2A - "cosec"^2A = (2sin^2A - 1)/(sin^2A *cos^2A)`.
If cot θ = `40/9`, find the values of cosec θ and sinθ,
We have, 1 + cot2θ = cosec2θ
1 + `square` = cosec2θ
1 + `square` = cosec2θ
`(square + square)/square` = cosec2θ
`square/square` = cosec2θ ......[Taking root on the both side]
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and sin θ = `1/("cosec" θ)`
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∴ sin θ = `9/41`
The value is cosec θ = `41/9`, and sin θ = `9/41`
