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प्रश्न
Prove the following identity :
`(sec^2θ - sin^2θ)/tan^2θ = cosec^2θ - cos^2θ`
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उत्तर
LHS = `(sec^2θ - sin^2θ)/tan^2θ`
= `(1/cos^2θ - sin^2θ)/(sin^2θ/cos^2θ)`
= `(1 - sin^2θcos^2θ)/((cos^2θ)/(sin^2θ/cos^2θ))`
= `(1 - sin^2θcos^2θ)/sin^2θ`
= `1/sin^2θ - (sin^2θcos^2θ)/(sin^2θ)`
= `cosec^2θ - cos^2θ`
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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`
