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Question
Prove the following identities:
`(1 - 2sin^2A)^2/(cos^4A - sin^4A) = 2cos^2A - 1`
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Solution
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.
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