Advertisements
Advertisements
प्रश्न
Find x , if `cos(2x - 6) = cos^2 30^circ - cos^2 60^circ`
Advertisements
उत्तर
`cos(2x - 6) = cos^2 30^circ - cos^2 60^circ`
⇒ `cos(2x - 6) = cos^2 (90^circ - 60^circ) - cos^2 60^circ`
⇒ `cos(2x - 6) = sin^2 60^circ - cos^2 60^circ`
⇒ `cos(2x - 6) = 1 - 2cos^2 60^circ = 1 - 2(1/2)^2 = 1 - 1/2 = 1/2`
⇒ `cos(2x - 6) = 1/2`
⇒ `cos(2x - 6) = cos60^circ`
⇒ `(2x - 6) = 60^circ`
⇒ `2x = 66^circ`
⇒ `x = 33^circ`
APPEARS IN
संबंधित प्रश्न
Prove that:
sec2θ + cosec2θ = sec2θ x cosec2θ
Prove that `cosA/(1+sinA) + tan A = secA`
As observed from the top of an 80 m tall lighthouse, the angles of depression of two ships on the same side of the lighthouse of the horizontal line with its base are 30° and 40° respectively. Find the distance between the two ships. Give your answer correct to the nearest meter.
Prove the following trigonometric identities.
tan2 θ − sin2 θ = tan2 θ sin2 θ
If a cos θ + b sin θ = 4 and a sin θ − b sin θ = 3, then a2 + b2 =
Prove the following identity :
`(1 + sinA)/(1 - sinA) = (cosecA + 1)/(cosecA - 1)`
Prove that ( 1 + tan A)2 + (1 - tan A)2 = 2 sec2A
cos θ . sec θ = ?
If `sqrt(3) tan θ` = 1, then find the value of sin2θ – cos2θ.
Eliminate θ if x = r cosθ and y = r sinθ.
