Advertisements
Advertisements
प्रश्न
If 24cosθ = 7 sinθ, find sinθ + cosθ.
उत्तर
24cosθ = 7sinθ
⇒ `("sin" θ)/("cos"θ) = (24)/(7)`
⇒ tanθ = `(24)/(7) = "Perpendicular"/"Base"`
Hypotenuse
= `sqrt(("Perpendicular")^2 + ("Base")^2`
= `sqrt((24)^2 + (7)^2`
= `sqrt(576 + 49)`
= `sqrt(625)`
= 25
sinθ + cosθ
= `"Perpendicular"/"Hypotenuse" + "Base"/"Hypotenuse"`
= `(24)/(25) + (7)/(25)`
= `(24 + 7)/(25)`
= `(31)/(25)`.
APPEARS IN
संबंधित प्रश्न
In the given figure, AD is perpendicular to BC. Find:
`(3)/("sin" x) + (4)/("cos" y) - 4 "tan" y`
If 4 sinθ = 3 cosθ, find `(6sinθ - 2cosθ )/(6sinθ + 2cosθ )`
If 3cosθ - 4sinθ = 2cosθ + sinθ, find tanθ.
If 4sinθ = `sqrt(13)`, find the value of 4sin3θ - 3sinθ
If 5tanθ = 12, find the value of `(2sinθ - 3cosθ)/(4sinθ - 9cosθ)`.
If 35 sec θ = 37, find the value of sin θ - sin θ tan θ.
If cotθ = `(1)/sqrt(3)`, show that `(1 - cos^2θ)/(2 - sin^2θ) = (3)/(5)`
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 = `(17)/(8)`, verify that `(3 - 4sin^2 "A")/(4 cos^2 "A" - 3)= (3 - tan^2"A")/(1 - 3tan^2"A")`
