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प्रश्न
Prove that sec2θ – cos2θ = tan2θ + sin2θ.
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उत्तर
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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Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
