Question

(b) Prove that :
sinA - cosA + 1 / sinA + cosA - 1 = 1 / secA - tanA
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Answer 1

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Answer 2

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