Advertisements
Advertisements
Question
Prove the following: `(tan(90° - θ)cotθ)/("cosec"^2 θ)` = cos2θ
Advertisements
Solution
L.H.S.
= `(tan(90° - θ)cotθ)/("cosec"^2 θ)`
= `(cotθ xx cotθ)/("cosec^2 θ)`
= `(cot^2 θ)/("cosec"^2 θ)`
= `((cos^2 θ)/(sin^2 θ))/((1)/(sin^2 θ)`
= cos2θ
= R.H.S.
APPEARS IN
RELATED QUESTIONS
Find the magnitude of angle A, if 2 sin A cos A - cos A - 2 sin A + 1 = 0
State for any acute angle θ whether tan θ increases or decreases as θ decreases.
If 3 tan A - 5 cos B = `sqrt3` and B = 90°, find the value of A
Solve for x : sin2 60° + cos2 (3x- 9°) = 1
Find the value of 'A', if (2 - cosec 2A) cos 3A = 0
Evaluate the following: `((sin3θ - 2sin4θ))/((cos3θ - 2cos4θ))` when 2θ = 30°
If θ = 15°, find the value of: cos3θ - sin6θ + 3sin(5θ + 15°) - 2 tan23θ
If `sqrt(3)` sec 2θ = 2 and θ< 90°, find the value of
cos2 (30° + θ) + sin2 (45° - θ)
In a rectangle ABCD, AB = 20cm, ∠BAC = 60°, calculate side BC and diagonals AC and BD.
If sec2θ = cosec3θ, find the value of θ if it is known that both 2θ and 3θ are acute angles.
