Advertisements
Advertisements
Question
If 4 sinθ = 3 cosθ, find tan2θ + cot2θ
Solution
4 sinθ = 3 cosθ
⇒ `(sin θ)/(cos θ) = (3)/(4)`
⇒ tan θ = `(sin θ)/(cos θ) = (3)/(4)`
⇒ cot θ = `(1)/(tanθ) = (4)/(3)`
tan2θ + cot2θ =
= `(3/4)^2 + (4/3)^2`
= `(9)/(16) + (16)/(9)`
= `(81 + 256)/(144)`
= `(337)/(144)`.
APPEARS IN
RELATED QUESTIONS
In tan θ = 1, find the value of 5cot2θ + sin2θ - 1.
In the given figure, ∠Q = 90°, PS is a median om QR from P, and RT divides PQ in the ratio 1 : 2. Find: `("tan" ∠"TSQ")/("tan"∠"PRQ")`
If 4 sinθ = 3 cosθ, find `(6sinθ - 2cosθ )/(6sinθ + 2cosθ )`
If 8tanA = 15, find sinA - cosA.
If 4sinθ = `sqrt(13)`, find the value of `(4sinθ - 3cosθ)/(2sinθ + 6cosθ)`
If a cotθ = b, prove that `("a"sinθ - "b"cosθ)/("a"sinθ + "b"cosθ) = ("a"^2 - "b"^2)/("a"^2 + "b"^2)`
If 12cosecθ = 13, find the value of `(sin^2θ - cos^2θ) /(2sinθ cosθ) xx (1)/tan^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 secA = `(17)/(8)`, verify that `(3 - 4sin^2 "A")/(4 cos^2 "A" - 3)= (3 - tan^2"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)`
