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: 5 cos x
In the given figure, AD is perpendicular to BC. Find: 15 tan y
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 2 tan ∠BAC - sin ∠BCD
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 35 sec θ = 37, find the value of sin θ - sin θ tan θ.
If cosecθ = `1(9)/(20)`, show that `(1 - sinθ + cosθ)/(1 + sinθ + cosθ) = (3)/(7)`
If b tanθ = a, find the values of `(cosθ + sinθ)/(cosθ - sinθ)`.
If a cotθ = b, prove that `("a"sinθ - "b"cosθ)/("a"sinθ + "b"cosθ) = ("a"^2 - "b"^2)/("a"^2 + "b"^2)`
If 12 cotθ = 13, find the value of `(2sinθ cosθ)/(cos^2θ - sin^2θ)`.
