Advertisements
Advertisements
प्रश्न
Prove that sin6A + cos6A = 1 – 3sin2A . cos2A.
Advertisements
उत्तर
L.H.S. = sin6A + cos6A
= (sin2A)3 + (cos2A)3
= (1 – cos2A)3 + (cos2A)3 ...`[(∵ sin^2A + cos^2A = 1),(∴ 1 - cos^2A = sin^2A)]`
= 1 – 3cos2A + 3(cos2A)2 – (cos2A)3 + cos6A ...[∵ (a – b)3 = a3 – 3a2b + 3ab2 – b3]
= 1 – 3 cos2A (1 – cos2A) – cos6A + cos6A
= 1 – 3 cos2A sin2A
= R.H.S.
∴ sin6A + cos6A = 1 – 3sin2A . cos2A
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identity.
`cos^2 A + 1/(1 + cot^2 A) = 1`
Prove the following trigonometric identities.
`(1 - cos theta)/sin theta = sin theta/(1 + cos theta)`
Prove the following trigonometric identities.
sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1
Prove the following trigonometric identities. `(1 - cos A)/(1 + cos A) = (cot A - cosec A)^2`
Prove that:
`cot^2A/(cosecA - 1) - 1 = cosecA`
`(1+ cos theta)(1- costheta )(1+cos^2 theta)=1`
Write the value of `(1 + tan^2 theta ) cos^2 theta`.
Write the value of `(1 + cot^2 theta ) sin^2 theta`.
What is the value of \[\sin^2 \theta + \frac{1}{1 + \tan^2 \theta}\]
Prove the following identity :
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (cosA + 1)/sinA`
Find the value of `θ(0^circ < θ < 90^circ)` if :
`tan35^circ cot(90^circ - θ) = 1`
If sec θ + tan θ = m, show that `(m^2 - 1)/(m^2 + 1) = sin theta`
Prove that `(sec θ - 1)/(sec θ + 1) = ((sin θ)/(1 + cos θ ))^2`
If A = 30°, verify that `sin 2A = (2 tan A)/(1 + tan^2 A)`.
If `sec θ = 41/40`, then find values of sin θ, cot θ, cosec θ.
If tan θ = 3, then `(4 sin theta - cos theta)/(4 sin theta + cos theta)` is equal to ______.
If 4 tanβ = 3, then `(4sinbeta-3cosbeta)/(4sinbeta+3cosbeta)=` ______.
If sinA + sin2A = 1, then the value of the expression (cos2A + cos4A) is ______.
If sinθ = `11/61`, then find the value of cosθ using the trigonometric identity.
Prove the following that:
`tan^3θ/(1 + tan^2θ) + cot^3θ/(1 + cot^2θ)` = secθ cosecθ – 2 sinθ cosθ
