Advertisements
Advertisements
प्रश्न
If 12cosecθ = 13, find the value of `(sin^2θ - cos^2θ) /(2sinθ cosθ) xx (1)/tan^2θ`.
उत्तर
12cosecθ = 13
⇒ cosecθ = `(13)/(12)`
⇒ sinθ = `(12)/(13) = "Perpendicular"/"Hypotenuse"`
⇒ Base
= `sqrt(("Hypotenuse")^2 - ("Perpendicular")^2`
= `sqrt((13)^2 - (12)^2`
= `sqrt(169 - 144)`
= `sqrt(25)`
= 5
cosθ = `"Base"/"Hypotenuse" = (5)/(13)`
tanθ= `"Perpendicular"/"Base" = (12)/(5)`
Now, `(sin^2θ - cos^2θ) /(2sinθ cosθ) xx (1)/tan^2θ`
= `((12/13)^2 - (5/13)^2)/(2(12/13)(5/13)) xx (1)/(12/5)^2`
= `(144/169 - 25/169)/(120/169) xx (25)/(144)`
= `(119)/(120) xx (25)/(144)`
= `(595)/(3456)`.
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: 15 tan y
In the given figure, AD is perpendicular to BC. Find: 5 cos x - 12 sin y + tan x
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
If 35 sec θ = 37, find the value of sin θ - sin θ tan θ.
If b tanθ = a, find the values of `(cosθ + sinθ)/(cosθ - sinθ)`.
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")`
If 3 tanθ = 4, prove that `sqrt(secθ - "cosec"θ)/(sqrt(secθ - "cosec"θ)) = (1)/sqrt(7)`.
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)`
