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प्रश्न
Prove the following identity :
`(cot^2θ(secθ - 1))/((1 + sinθ)) = sec^2θ((1-sinθ)/(1 + secθ))`
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उत्तर
LHS = `(cot^2θ(secθ - 1))/((1 + sinθ)) `
= `(cot^2θ(secθ - 1)(1 - sinθ)(secθ + 1))/((1 + sinθ)(1 - sinθ)(secθ + 1))`
= `(cot^2θ(secθ - 1)(secθ + 1)(1 - sinθ))/((1 + sinθ)(1 - sinθ)(secθ + 1))`
= `(cot^2θ(sec^2θ - 1)(1 - sinθ))/((1 - sin^2θ)(1 + secθ))`
= `(cot^2θ(tan^2θ)(1 - sinθ))/((cos^2θ)(1 + secθ))` (∵ `tan^2θ = sec^2θ - 1,1 - sin^2θ = cos^2θ`)
= `((cotθtanθ)^2(1 - sinθ))/((cos^2θ)(1 + secθ))`
= `(1(1 - sinθ))/((cos^2θ)(1 + secθ))` (∵ cotθtanθ = 1)
= `sec^2θ((1 - sinθ)/(1 + secθ))`
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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]
cosec θ = `41/9`
and sin θ = `1/("cosec" θ)`
sin θ = `1/square`
∴ sin θ = `9/41`
The value is cosec θ = `41/9`, and sin θ = `9/41`
