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Question
Prove that `(sin "A" - cos "A" + 1)/(sin "A" + cos "A" - 1) = 1/(sec "A" - tan "A")`
Solution
L.H.S = `(sin"A" - cos "A" + 1)/(sin "A" + cos "A" - 1)`
`= (tan "A" -1 + sec"A")/(tan "A" + 1 - sec "A")` [Dividing numerator & denominator by cos A]
`= ((tan "A" + sec "A") -1)/((tan "A" - sec "A") + 1)`
`= ((tan "A" + sec"A") - 1(tan"A" - sec "A"))/((tan "A" + sec"A") + 1(tan"A" - sec "A"))`
`= ((tan^2 "A" + sec^2 "A") - (tan "A" - sec "A"))/ ((tan "A" + sec "A" + 1) - (tan "A" - sec "A"))`
`= (-1 - tan "A" + sec "A")/((tan "A" - sec "A" + 1)(tan "A" - sec "A"))`
`= -1/(tan "A" - sec "A")`
`= 1/(sec "A" - tan "A")`
LHS = RHS
Hence proved.
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