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Question
Prove that `(1 + sec "A")/"sec A" = (sin^2"A")/(1 - cos"A")`
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Solution
L.H.S = `(1 + sec "A")/"sec A"`
= `1/"sec A" + "sec A"/"sec A"`
= cos A + 1
= `(1 + cos "A") xx (1 - cos"A")/(1 - cos"A")`
= `(1 - cos^2"A")/(1 - cos"A")`
= `(sin^2"A")/(1 - cos"A")` .......`[(because sin^2"A" + cos^2"A" = 1),(therefore 1 - cos^2"A" = sin^2"A")]`
= R.H.S
∴ `(1 + sec "A")/"sec A" = (sin^2"A")/(1 - cos"A")`
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