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प्रश्न
Prove the following identity :
`(1 + sinθ)/(cosecθ - cotθ) - (1 - sinθ)/(cosecθ + cotθ) = 2(1 + cotθ)`
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उत्तर
LHS = `(1 + sinθ)/(cosecθ - cotθ) - (1 - sinθ)/(cosecθ + cotθ)`
= `((1 + sinθ)(cosecθ + cotθ) - (1 - sinθ)(cosecθ - cotθ))/(cosec^2θ - cot^2θ)`
= `(cosecθ + cotθ + 1 + cosθ - cosecθ + cotθ + 1 - cosθ)/(1 + cot^2θ - cot^2θ)` (∵ `cosec^2θ = 1 + cot^2θ`)
= 2 + 2cotθ = 2(1 + cotθ)
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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)`
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∴ sin θ = `9/41`
The value is cosec θ = `41/9`, and sin θ = `9/41`
