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Question
If sinθ = `(3)/(4)`, prove that `sqrt(("cosec"^2θ - cot^2θ)/(sec^2θ - 1)) = sqrt(7)/(3)`.
Solution
sinθ = `(3)/(4)`
`"Perpendicular"/"Hypotenuse" = (3)/(4)`
Base
= `sqrt(("Hypotenuse")^2 - ("Perpendicuar")^2`
= `sqrt(16 - 9)`
= `sqrt(7)`
cosecθ = `(4)/(3)`
cotθ = `"Base"/"Perpendicular" = sqrt(7)/(3)`
secθ = `"Hypotenuse"/"Base" = (4)/sqrt(7)`
Tp prove `sqrt(("cosec"^2θ - cot^2θ)/(sec^2θ - 1)) = sqrt(7)/(3)`
`sqrt(("cosec"^2θ - cot^2θ)/(sec^2θ - 1)`
= `sqrt(((4/3)^2 - (sqrt(7)/3)^2)/((4/sqrt(7))^2 - 1)`
= `sqrt((16 /9 - 7/9)/(16/7 - 1)`
= `sqrt(((16 - 7)/9)/((16 - 7)/7)`
= `sqrt((9/9)/(9/7)`
= `sqrt((1)/(9/7)`
= `sqrt(7/9)`
= `sqrt(7)/(3)`.
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