Advertisements
Advertisements
प्रश्न
If 3 tanθ = 4, prove that `sqrt(secθ - "cosec"θ)/(sqrt(secθ - "cosec"θ)) = (1)/sqrt(7)`.
उत्तर
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)`.
APPEARS IN
संबंधित प्रश्न
In tan θ = 1, find the value of 5cot2θ + sin2θ - 1.
In the given figure, AD is perpendicular to BC. Find: 5 cos x
In the given figure, AD is perpendicular to BC. Find:
`(3)/("sin" x) + (4)/("cos" y) - 4 "tan" y`
In a right-angled triangle ABC, ∠B = 90°, BD = 3, DC = 4, and AC = 13. A point D is inside the triangle such as ∠BDC = 90°.
Find the values of 3 - 2 cos ∠BAC + 3 cot ∠BCD
If 5cosθ = 3, find the value of `(4cosθ - sinθ)/(2cosθ + sinθ)`
If cotθ = `(1)/sqrt(3)`, show that `(1 - cos^2θ)/(2 - sin^2θ) = (3)/(5)`
If cosecθ = `1(9)/(20)`, show that `(1 - sinθ + cosθ)/(1 + sinθ + cosθ) = (3)/(7)`
If 12cosecθ = 13, find the value of `(sin^2θ - cos^2θ) /(2sinθ cosθ) xx (1)/tan^2θ`.
If secA = `(5)/(4)`, cerify that `(3sin"A" - 4sin^3"A")/(4cos^3"A" - 3cos"A") = (3tan"A" - tan^3"A")/(1 - 3tan^2"A")`.
If tan θ = `"m"/"n"`, show that `"m sin θ - n cos θ"/"m sinθ + n cos θ" = ("m"^2 - "n"^2)/("m"^2 + "n"^2)`
