Advertisements
Advertisements
प्रश्न
If b tanθ = a, find the values of `(cosθ + sinθ)/(cosθ - sinθ)`.
उत्तर
b tanθ = a
⇒ tanθ = `"a"/"b"`
Consider `(cosθ + sinθ)/(cosθ - sinθ)`
Dividing the numerator and demoninator by cosθ, we get
`(cosθ + sinθ)/(cosθ - sinθ)`
= `(1 + sinθ/cosθ)/(1 - sinθ/cosθ)`
= `(1 + tanθ)/(1 - tanθ)`
= `(1 + "a"/"b")/(1 - "a"/"b")`
= `(("b" + "a")/"b")/(("b" - "a")/"b")`
= `(("b" + "a"))/(("b" - "a")`.
APPEARS IN
संबंधित प्रश्न
In the given figure, AD is perpendicular to BC. Find: 15 tan y
In the given figure, AD is perpendicular to BC. Find: 5 cos x - 12 sin y + tan x
In the given figure, AD is perpendicular to BC. Find:
`(3)/("sin" x) + (4)/("cos" y) - 4 "tan" y`
If 24cosθ = 7 sinθ, find sinθ + cosθ.
If 3cosθ - 4sinθ = 2cosθ + sinθ, find tanθ.
If 5tanθ = 12, find the value of `(2sinθ - 3cosθ)/(4sinθ - 9cosθ)`.
If cotθ = `(1)/sqrt(3)`, show that `(1 - cos^2θ)/(2 - sin^2θ) = (3)/(5)`
If 12cosecθ = 13, find the value of `(sin^2θ - cos^2θ) /(2sinθ cosθ) xx (1)/tan^2θ`.
If 12 cotθ = 13, find the value of `(2sinθ cosθ)/(cos^2θ - sin^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")`.
