Advertisements
Advertisements
प्रश्न
Prove that: 2(sin6θ + cos6θ) - 3 ( sin4θ + cos4θ) + 1 = 0.
Advertisements
उत्तर
LHS = 2(sin6θ + cos6θ) - 3 ( sin4θ + cos4θ) + 1
= 2( sin2θ + cos2θ ) [ sin4θ + cos4θ - sin2θ.cos2θ ] - 3[ ( sin2θ + cos2θ )2 - 2sin2θ. cos2θ + 1
= 2 x 1 [ ( sin2θ + cos2θ )2 - 2 sin2θ.cos2θ - sin2θ.cos2θ ] - 3[ (1)2 - 2sin2θ. cos2θ ] + 1
= 2 [ (1)2 - 3 sin2θ.cos2θ ] - 3 [ 1 - 2 sin2θ. cos2θ ] + 1
= 2 - 6 sin2θ. cos2θ - 3 + 6 sin2θ. cos2θ + 1
= - 1 + 1 = 0
= RHS
Hence proved.
संबंधित प्रश्न
If tanθ + sinθ = m and tanθ – sinθ = n, show that `m^2 – n^2 = 4\sqrt{mn}.`
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 trigonometric identities:
`(1 + tan^2 theta)/(1 + cot^2 theta) = ((1 - tan theta)/(1 - cot theta))^2 = tan^2 theta`
`Prove the following trigonometric identities.
`(sec A - tan A)^2 = (1 - sin A)/(1 + sin A)`
Prove the following identities:
`(sintheta - 2sin^3theta)/(2cos^3theta - costheta) = tantheta`
Prove that:
(tan A + cot A) (cosec A – sin A) (sec A – cos A) = 1
`(sectheta- tan theta)/(sec theta + tan theta) = ( cos ^2 theta)/( (1+ sin theta)^2)`
Prove the following identity:
tan2A − sin2A = tan2A · sin2A
Prove the following identity :
(secA - cosA)(secA + cosA) = `sin^2A + tan^2A`
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= `square (1 - (sin^2θ)/(tan^2θ))`
= `tan^2θ (1 - square/((sin^2θ)/(cos^2θ)))`
= `tan^2θ (1 - (sin^2θ)/1 xx (cos^2θ)/square)`
= `tan^2θ (1 - square)`
= `tan^2θ xx square` ...[1 – cos2θ = sin2θ]
= R.H.S.
