Advertisements
Advertisements
प्रश्न
If cotθ = `sqrt(7)`, show that `("cosec"^2θ -sec^2θ)/("cosec"^2θ + sec^2θ) = (3)/(4)`
उत्तर
cotθ = `sqrt(7)`
⇒ `cosθ/sinθ = sqrt(7)`
⇒ `"Base"/"Hypotenuse" xx "Hypotenuse"/"Perpendicular" = sqrt(7)/(1)`
⇒ `"Base"/"Perpendicular" = sqrt(7)/(1)`
Hypotenuse
= `sqrt(("Perpendicular")^2 + ("Base")^2`
= `sqrt(1 + 7)`
= `2sqrt(2)`
To show: `("cosec"^2θ -sec^2θ)/("cosec"^2θ + sec^2θ) = (3)/(4)`
`("cosec"^2θ -sec^2θ)/("cosec"^2θ + sec^2θ)`
= `(("Hypotenuse"/"Perpendicular")^2 - ("Hypotenuse"/"Base")^2)/(("Hypotenuse"/"Perpendicular")^2 + ("Hypotenuse"/"Base")^2)`
= `((2sqrt(2)/1)^2 - (2sqrt(2)/sqrt(7))^2)/(((2sqrt(2))/1)^2 + (2sqrt(2)/sqrt(7))^2`
= `(8/1 - 8/7)/(8/1 + 8/7)`
= `((56 - 8)/7)/((56 + 8)/7)`
= `(48)/(64)`
= `(3)/(4)`.
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:
`(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 4 sinθ = 3 cosθ, find `(6sinθ - 2cosθ )/(6sinθ + 2cosθ )`
If 5cosθ = 3, find the value of `(4cosθ - sinθ)/(2cosθ + sinθ)`
If 4sinθ = `sqrt(13)`, find the value of `(4sinθ - 3cosθ)/(2sinθ + 6cosθ)`
If a cotθ = b, prove that `("a"sinθ - "b"cosθ)/("a"sinθ + "b"cosθ) = ("a"^2 - "b"^2)/("a"^2 + "b"^2)`
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 sinθ = `(3)/(4)`, prove that `sqrt(("cosec"^2θ - cot^2θ)/(sec^2θ - 1)) = sqrt(7)/(3)`.
