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 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: 15 tan y
If 4sinθ = `sqrt(13)`, find the value of `(4sinθ - 3cosθ)/(2sinθ + 6cosθ)`
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 cotθ = `sqrt(7)`, show that `("cosec"^2θ -sec^2θ)/("cosec"^2θ + sec^2θ) = (3)/(4)`
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")`.
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)`
