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प्रश्न
Prove the following identities:
`(1 - 2sin^2A)^2/(cos^4A - sin^4A) = 2cos^2A - 1`
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उत्तर
L.H.S. `(1 - 2sin^2A)^2/(cos^4A - sin^4A)`
= `(1 - 2sin^2A)^2/((cos^2A)^2 - (sin^2A)^2`
= `(1 - 2sin^2A)^2/((cos^2A + sin^2A)(cos^2A - sin^2A))`
= `(1 - 2sin^2A)^2/((1)(1 - sin^2A - sin^2A)` ...[∵ cos2 A = 1 – sin2 A]
= `(1 - 2sin^2A)^2/((1 - 2sin^2A))`
= 1 – 2 sin2 A
= 1 – 2 (1 – cos2 A)
= 1 – 2 + 2 cos2 A
= 2 cos2 A – 1 = R.H.S.
Hence the result is proved.
संबंधित प्रश्न
Prove that `(sin theta)/(1-cottheta) + (cos theta)/(1 - tan theta) = cos theta + sin theta`
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`sinA/(1 + cosA) = cosec A - cot A`
`(1+ tan^2 theta)/(1+ tan^2 theta)= (cos^2 theta - sin^2 theta)`
Prove the following identity :
`(1 - tanA)^2 + (1 + tanA)^2 = 2sec^2A`
Prove the following identity :
`tan^2A - tan^2B = (sin^2A - sin^2B)/(cos^2Acos^2B)`
Prove the following identity :
`(cot^2θ(secθ - 1))/((1 + sinθ)) = sec^2θ((1-sinθ)/(1 + secθ))`
Prove the following identities.
cot θ + tan θ = sec θ cosec θ
Choose the correct alternative:
sec2θ – tan2θ =?
Prove the following:
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