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