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Question
sec4A(1 - sin4A) - 2tan2A = 1
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Solution
डावी बाजू = sec4A(1 - sin4A) - 2tan2A
= sec4A[12 – (sin2A)2] – 2tan2A
= sec4A .(1 – sin2A) (1 + sin2A) – 2tan2A
= sec4A cos2A (1 + sin2A) – 2tan2A ...`[(∵ sin^2θ + cos^2θ = 1), (∴ 1 - sin^2θ = cos^2θ)]`
= `1/cos^4A . cos^2A(1 + sin^2A) - 2tan^2A`
= `1/cos^2A (1 + sin^2A) - 2tan^2A`
= `1/cos^2A + sin^2A/cos^2A - 2tan^2A`
= sec2A + tan2A – 2tan2A
= sec2A – tan2A
= 1 ................[∵ sec2θ – tan2θ = 1]
= उजवी बाजू
∴ sec4A(1 - sin4A) - 2tan2A = 1
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