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Question
Prove that `(tanA + secA - 1)/(tanA - secA + 1) = (1 + sinA)/cosA`
Solution
L.H.S. `(tanA + secA - 1)/(tanA - secA + 1)`
= `(tan A + (secA - 1))/(tanA - (secA - 1))`
= `([tan A + (secA - 1)][tanA + (secA - 1)])/([tanA - (sec A - 1)][tanA + (secA - 1)]` ......[Rationalizing the denominator]
= `[tan A + (sec A - 1)]^2/(tan^2A - (secA - 1)^2`
= `(tan^2A + (secA - 1)^2 + 2tanA(secA - 1))/(tan^2A - (sec^2A + 1 - 2secA))`
= `(tan^2 + sec^2A + 1 - 2secA + 2tanA secA - 2tanA)/(tan^2A - sec^2A - 1 + 2secA)`
= `(sec^2A + sec^2A - 2secA + 2tanAsecA - 2tanA)/(-1 - 1 + 2secA)`
= `(2sec^2A - 2secA - 2secA + 2tanAsecA - 2tanA)/(2secA - 2)`
= `(sec^2A - secA + secA + secA tanA - tanA)/(secA - 1)`
= `(secA(secA - 1) + tanA(secA- 1))/(secA- 1)`
= `((secA - 1)(secA+ tanA))/((secA - 1))`
= sec A + tan A
= `1/cosA + sinA/cosA`
= `(1 + sinA)/cosA`
= R.H.S.
Hence proved.
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