Advertisements
Advertisements
Question
Prove that sin (90° - θ) cos (90° - θ) = tan θ. cos2θ.
Advertisements
Solution
LHS = sin (90° - θ) cos (90° - θ)
LHS = cos θ. sin θ
RHS = tan θ. cos2θ
RHS = `sin θ/cos θ` x cos2θ
RHS = cos θ. sin θ
∴ LHS = RHS
Hence proved.
RELATED QUESTIONS
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
`(sintheta - 2sin^3theta)/(2costheta - costheta) =tan theta`
Prove the following identities:
`(cotA - cosecA)^2 = (1 - cosA)/(1 + cosA)`
Prove that:
`cot^2A/(cosecA - 1) - 1 = cosecA`
Write the value of `( 1- sin ^2 theta ) sec^2 theta.`
Write the value of `3 cot^2 theta - 3 cosec^2 theta.`
Prove that `(sec θ - 1)/(sec θ + 1) = ((sin θ)/(1 + cos θ ))^2`
Prove the following identities.
`(1 - tan^2theta)/(cot^2 theta - 1)` = tan2 θ
Prove the following:
`tanA/(1 + sec A) - tanA/(1 - sec A)` = 2cosec A
Prove that `(1 + sec theta - tan theta)/(1 + sec theta + tan theta) = (1 - sin theta)/cos theta`
If 2 cos θ + sin θ = `1(θ ≠ π/2)`, then 7 cos θ + 6 sin θ is equal to ______.
