Advertisements
Advertisements
प्रश्न
Prove that sin4θ - cos4θ = sin2θ - cos2θ
= 2sin2θ - 1
= 1 - 2 cos2θ
Advertisements
उत्तर
L.H.S. = sin4θ - cos4θ
L.H.S. = (sin2θ)2 - (cos2θ)2
L.H.S. = (sin2θ - cos2θ)(sin2θ + cos2θ)
L.H.S. = (sin2θ - cos2θ) x 1
L.H.S. = sin2θ - cos2θ
L.H.S. = R.H.S.
L.H.S.= sin2θ - (1 - sin2θ)
L.H.S. = sin2θ - 1 + sin2θ
L.H.S. = 2sin2θ - 1
L.H.S. = R.H.S
L.H.S. = 2(1 - cos2θ) - 1
L.H.S. = 2 - 2cos2θ - 1
L.H.S. = 1 - 2cos2θ
L.H.S. = R.H.S.
संबंधित प्रश्न
Prove the following trigonometric identities.
`sqrt((1 - cos A)/(1 + cos A)) = cosec A - cot A`
if `a cos^3 theta + 3a cos theta sin^2 theta = m, a sin^3 theta + 3 a cos^2 theta sin theta = n`Prove that `(m + n)^(2/3) + (m - n)^(2/3)`
`(sec theta -1 )/( sec theta +1) = ( sin ^2 theta)/( (1+ cos theta )^2)`
Prove the following identity :
`(1 + tan^2A) + (1 + 1/tan^2A) = 1/(sin^2A - sin^4A)`
Express (sin 67° + cos 75°) in terms of trigonometric ratios of the angle between 0° and 45°.
There are two poles, one each on either bank of a river just opposite to each other. One pole is 60 m high. From the top of this pole, the angle of depression of the top and foot of the other pole are 30° and 60° respectively. Find the width of the river and height of the other pole.
Prove that sin2 θ + cos4 θ = cos2 θ + sin4 θ.
Prove that:
`(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A) = 2/(2 sin^2 A - 1)`
If 3 sin A + 5 cos A = 5, then show that 5 sin A – 3 cos A = ± 3.
If sinθ = `11/61`, then find the value of cosθ using the trigonometric identity.
