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प्रश्न
Prove the following identity :
`sec^4A - sec^2A = sin^2A/cos^4A`
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उत्तर
`sec^4A - sec^2A = 1/cos^4A - 1/cos^2A`
= `(1 - cos^2A)/cos^4A`
= `sin^2A/cos^4A` [∵ `sin^2A = 1 - cos^2A`]
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संबंधित प्रश्न
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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`
