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Question
Prove that sec2θ – cos2θ = tan2θ + sin2θ.
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Solution
L.H.S. = sec2θ – cos2θ
= sec2θ – (1 – sin2θ) ...`[(∵ sin^2θ + cos^2θ = 1),(∴ 1 - sin^2θ = cos^2θ)]`
= sec2θ – 1 + sin2θ
= tan2θ + sin2θ ...`[(∵ 1 + tan^2θ = sec^2θ),(∴ tan^2θ = sec^2θ - 1)]`
= R.H.S.
∴ sec2θ – cos2θ = tan2θ + sin2θ
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