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प्रश्न
Prove that: 2(sin6θ + cos6θ) - 3 ( sin4θ + cos4θ) + 1 = 0.
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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
Prove the following trigonometric identities.
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`sin theta / ((1+costheta))+((1+costheta))/sin theta=2cosectheta`
Prove the following identities:
`(1 + cos theta - sin^2 theta )/(sin theta (1 + cos theta)) = cot theta`
Prove the following identities:
`(1 + cos theta + sin theta)/(1 + cos theta - sin theta) = (1 + sin theta )/(cos theta)`
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If sin θ + cos θ = p and sec θ + cosec θ = q, then prove that q(p2 – 1) = 2p.
