Question

prove that (sinA-cosA+1)/(sinA+cosA-1)=1/(secA-tanA)​

Answer 1

Answer:

LHS = sinA-cosA+1/sinA+cosA-1

divide both numerator and denominator by cosA

LHS=(tanA−1+secA)/(tanA+1−secA)

Now

sec2A=1+tan2A

sec2A−tan2A=1

Using above relation at denominator of LHS

LHS=(tanA−1+secA)/(tanA−secA+sec2A−tan2A)

LHS=(tanA−1+secA)/((secA−tanA)(−1+secA+tanA))

LHS=1/(secA−tanA)

LHS=RHS

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Answer 2

$\frac{sinA - cosA + 1}{sinA + cosA - 1} \\ \\ = \frac{ \frac{sinA - cosA + 1}{cosA} }{ \frac{sinA + cosA - 1}{cosA} } \\ \\ = \frac{tanA + secA - 1}{tanaA- secA + 1} \\ \\ = \frac{tanA + secA - 1}{( {sec}^{2} A - {tan}^{2}A) - (secA - tanA) } \\ \\ = \frac{tanA + secA - 1}{(secA - tanA)(secA + tanA - 1)} \\ \\ = \frac{1}{secA - tanA}$