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Question
सिद्ध कीजिए।
sec4A (1 - sin4A) - 2tan2A = 1
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Solution
बायाँ पक्ष = sec4A (1 - sin4A) - 2tan2A
= `sec^4A - sec^4A.sin^4A - 2tan^2A`
= `1/cos^4A - 1/cos^4A.sin^4A - 2sin^2A/cos^2A` ....................`(∵ secA = 1/cosA, tanA = sinA/cosA)`
= `1/cos^4A - sin^4A/cos^4A - (2sin^2A)/cos^2A`
= `((1 - sin^4A))/cos^4A - (2sin^2A)/cos^2A`
= `[[1 - (sin^2A)^2]]/cos^4A - (2sin^2A)/cos^2A`
= `((1 + sin^2A)(1 - sin^2A))/cos^4A- (2sin^2A)/cos^2A`
= `((1 + sin^2A) xx cos^2A)/cos^4A - (2sin^2A)/cos^2A` ...............`((∵ sin^2A + cos^2A = 1),(∴ cos^2A = 1 - sin^2A))`
= `(1 + sin^2A)/cos^2A - (2sin^2A)/cos^2A`
= `(1 + sin^2A - 2sin^2A)/cos^2A`
= `(1 - sin^2A)/cos^2A`
= `cos^2A/cos^2A` ................`[(∵ sin^2A + cos^2A = 1),(∴ cos^2A = 1 - sin^2A)]`
= 1 = दायाँ पक्ष
∴ बायाँ पक्ष = दायाँ पक्ष
∴ sec4A (1 - sin4A) - 2tan2A = 1.
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