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प्रश्न
Prove that `sec^2A - "cosec"^2A = (2sin^2A - 1)/(sin^2A *cos^2A)`.
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उत्तर
L.H.S. = sec2A – cosec2A
= `1/(cos^2A) - 1/(sin^2A)`
= `(sin^2A - cos^2A)/(cos^2A*sin^2A)`
= `(sin^2A - (1 - sin^2A))/(sin^2A*cos^2A)` ...`[(∵ sin^2A + cos^2A = 1),(∴ 1 - sin^2A = cos^2A)]`
= `(sin^2A - 1 + sin^2A)/(sin^2A*cos^2A)`
= `(2sin^2"A" - 1)/(sin^2"A"*cos^2"A")`
= R.H.S.
∴ `sec^2A - "cosec"^2A = (2sin^2A - 1)/(sin^2A*cos^2A)`
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