Advertisements
Advertisements
प्रश्न
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)`
उत्तर
tan θ = `"m"/"n" = "Perpendicular"/"Base"`
Hypotenuse
= `sqrt(("Perpendicular")^2 + ("Base")^2`
= `sqrt("m"^2 + "n"^2`
sin θ = `("m"/sqrt("m"^2 + "n"^2))`
cos θ = `("n"/sqrt("m"^2 + "n"^2))`
To show: `"m sin θ - n cos θ"/"m sinθ + n cos θ" = ("m"^2 - "n"^2)/("m"^2 + "n"^2)`.
`"m sin θ - n cos θ"/"m sinθ + n cos θ"`
= `(("m"/sqrt("m"^2 + "n"^2)) - "n"("n"/sqrt("m"^2 + "n"^2)))/(("m"/sqrt("m"^2 + "n"^2)) + "n"("n"/sqrt("m"^2 + "n"^2))`
= `(("m"^2 - "n"^2)/(sqrt("m"^2 + "n"^2)))/(("m"^2 + "n"^2)/(sqrt("m"^2 + "n"^2)`
= `("m"^2 - "n"^2)/(sqrt("m"^2 + "n"^2)) xx sqrt("m"^2 + "n"^2)/("m"^2 + "n"^2)`
= `("m"^2 - "n"^2)/("m"^2 + "n"^2)`.
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`
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 24cosθ = 7 sinθ, find sinθ + cosθ.
If 8tanA = 15, find sinA - cosA.
If a cotθ = b, prove that `("a"sinθ - "b"cosθ)/("a"sinθ + "b"cosθ) = ("a"^2 - "b"^2)/("a"^2 + "b"^2)`
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 = `(17)/(8)`, verify that `(3 - 4sin^2 "A")/(4 cos^2 "A" - 3)= (3 - tan^2"A")/(1 - 3tan^2"A")`
