Question

If 3 tanθ = 4, prove that `sqrt(secθ - "cosec"θ)/(sqrt(secθ - "cosec"θ)) = (1)/sqrt(7)`.

Answer 1

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)`.