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