Question

Prove that `(sin "A" - cos "A" + 1)/(sin "A" + cos "A" - 1) = 1/(sec "A" - tan "A")`

Answer 1

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.