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प्रश्न
Prove that sin4θ - cos4θ = sin2θ - cos2θ
= 2sin2θ - 1
= 1 - 2 cos2θ
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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:
(i) (1 – sin2θ) sec2θ = 1
(ii) cos2θ (1 + tan2θ) = 1
Prove the following trigonometric identities.
`(1 - cos theta)/sin theta = sin theta/(1 + cos theta)`
Prove the following identities:
`(1 + sinA)/cosA + cosA/(1 + sinA) = 2secA`
Prove the following identities:
cosec4 A (1 – cos4 A) – 2 cot2 A = 1
` tan^2 theta - 1/( cos^2 theta )=-1`
` (sin theta + cos theta )/(sin theta - cos theta ) + ( sin theta - cos theta )/( sin theta + cos theta) = 2/ ((1- 2 cos^2 theta))`
If `sec theta + tan theta = x," find the value of " sec theta`
If a cot θ + b cosec θ = p and b cot θ − a cosec θ = q, then p2 − q2
`5/(sin^2theta) - 5cot^2theta`, complete the activity given below.
Activity:
`5/(sin^2theta) - 5cot^2theta`
= `square (1/(sin^2theta) - cot^2theta)`
= `5(square - cot^2theta) ......[1/(sin^2theta) = square]`
= 5(1)
= `square`
If sec θ = `41/40`, then find values of sin θ, cot θ, cosec θ
