Advertisements
Advertisements
Question
Prove the following: `(tan(90° - θ)cotθ)/("cosec"^2 θ)` = cos2θ
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
Calculate the value of A, if (sin A - 1) (2 cos A - 1) = 0
Solve for x : cos2 30° + sin2 2x = 1
Find the value of 'A', if cot 3A = 1
If sin α + cosβ = 1 and α= 90°, find the value of 'β'.
If θ = 30°, verify that: sin2θ = `(2tanθ)/(1 ++ tan^2θ)`
If A = B = 60°, verify that: sin(A - B) = sinA cosB - cosA sinB
In the given figure, AB and EC are parallel to each other. Sides AD and BC are 1.5 cm each and are perpendicular to AB. Given that ∠AED = 45° and ∠ACD = 30°. Find:
a. AB
b. AC
c. AE
Find:
a. BC
b. AD
c. AC
Find the value 'x', if:
Evaluate the following: cot20° cot40° cot45° cot50° cot70°
