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प्रश्न
Prove the following identities:
`(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1) = 2sec^2A`
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उत्तर
L.H.S. `=(cosec A)/(cosecA - 1) + (cosecA)/(cosecA + 1)`
= `(cosec A (cosec A + 1) + cosec A (cosec A - 1))/((cosec A - 1) (cosec A + 1))`
= `(cosec^2 A+cosec A + cosec^2 A-cosec A)/((cosec A)^2 - (1)^2)`
= `(2 cosec^2 A)/(cosec^2 A - 1)`
= `(2 cosec^2 A)/(cot^2 A)` ...(∵ cosec2 A – 1 = cot2 A)
= `2(1/cancel(sin^2A))/(cos^2A/cancel(sin^2A))`
= `2/cos^2A`
= 2 sec2 A
= R.H.S.
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If sin2 θ cos2 θ (1 + tan2 θ) (1 + cot2 θ) = λ, then find the value of λ.
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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`
