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Question
If 3 tanθ = 4, prove that `sqrt(secθ - "cosec"θ)/(sqrt(secθ - "cosec"θ)) = (1)/sqrt(7)`.
Solution
3 tanθ = 4
⇒ tanθ = `(4)/(3) = "Perpendicular"/"Base"`
Hypotenuse
= `sqrt(("Perpendicular")^2 + ("Base")^2`
= `sqrt((4)^2 + (3)^2`
= `sqrt(16 + 9)`
= `sqrt(25)`
= 5
secθ = `"Hypotenuse"/"Base" = (5)/(3)`
cosecθ = `"Hypotenuse"/"Perpendicular" = (5)/(4)`
To prove: `sqrt(secθ - "cosec"θ)/(sqrt(secθ - "cosec"θ)) = (1)/sqrt(7)`.
`sqrt(secθ - "cosec"θ)/(sqrt(secθ - "cosec"θ))`
= `(sqrt(5/3 - 5/4))/(sqrt(5/3 + 5/4)`
= `(sqrt(20 - 15)/12)/(sqrt(20 + 15)/12)`
= `(sqrt(5/12))/(sqrt(35/12)`
= `sqrt(5)/sqrt(12) xx sqrt(12)/(sqrt(35)`
= `sqrt(5)/sqrt(12) xx sqrt(12)/(sqrt(5) xx sqrt(7))`
= `(1)/sqrt(7)`.
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