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प्रश्न
Prove that sec2θ – cos2θ = tan2θ + sin2θ.
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उत्तर
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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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`
