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प्रश्न
Prove the following identity :
`(1 - cos^2θ)sec^2θ = tan^2θ`
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उत्तर
`(1 - cos^2θ)sec^2θ = tan^2θ`
Consider L.H.S = `sin^2θ1/cos^2θ`
= `tan^2θ` = RHS
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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`
