Advertisements
Advertisements
प्रश्न
If 8tanA = 15, find sinA - cosA.
उत्तर
8tanA = 15
⇒ tanA = `(15)/(8) = "Perpendicular"/"Base"`
Hypotenuse
= `sqrt(("Perpendicular")^2 + ("Base")^2`
= `sqrt((15)^2 + (8)2`
= `sqrt(225 + 64)`
= `sqrt(289)`
= 17
sinA - cosA = `"Perpendicular"/"Hypotenuse" - "Base"/"Hypotenuse"`
= `(15)/(17) - (8)/(17)`
= `(15 - 8)/(17)`
sinA - cosA = `(7)/(17)`.
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:
`(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 4 sinθ = 3 cosθ, find tan2θ + cot2θ
If 5cosθ = 3, find the value of `(4cosθ - sinθ)/(2cosθ + sinθ)`
If 5tanθ = 12, find the value of `(2sinθ - 3cosθ)/(4sinθ - 9cosθ)`.
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")`