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Question
Prove that (1 – cos2A) . sec2B + tan2B (1 – sin2A) = sin2A + tan2B.
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Solution
L.H.S. = (1 – cos2A) . sec2B + tan2B (1 – sin2A)
= `sin^2A * 1/(cos^2B) + (sin^2B)/(cos^2B) (1 - sin^2A)` ...`[(∵ sin^2A + cos^2A = 1),(∴ 1 - cos^2A = sin^2A)]`
= `(sin^2A)/(cos^2B) + (sin^2B)/(cos^2B) - (sin^2A sin^2B)/(cos^2B)`
= `(sin^2A)/(cos^2B) - (sin^2A sin^2B)/(cos^2B) + (sin^2B)/(cos^2B)`
= `(sin^2A)/(cos^2B) (1 - sin^2B) + tan^2B`
= `(sin^2A)/(cos^2B) (cos^2B) + tan^2B`
= sin2A + tan2B
= R.H.S.
∴ (1 – cos2A) . sec2B + tan2B (1 – sin2A) = sin2A + tan2B
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