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Answer 1

Answer:

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Step-by-step explanation:

$to \: find = \\ \\ \frac{sin \: A - cos \: A - 1}{sin \:A + cos \: A - 1} \\ \\ so \: conjugate \: of \: \: sin \: A + cos \: A - 1 \\ is \: \: sin \: A - cos \: A +1 \\ \\ so \: by \: conjugate \: method \\ we \: have \\ \\ = \frac{sin \:A - cos \:A - 1 }{sin \:A + cos \:A - 1 } \times \frac{sin \: A - cos \:A + 1 }{sin \: A - cos \: A + 1} \\ \\ we \: know \: that \\ \\ (a - b - c)(a - b + c) = a {}^{2} + b {}^{2} - c {}^{2} - 2ab \\ \\ (a + b - c)(a - b + c) = a {}^{2} - b {}^{2} - c {}^{2} + 2bc \\ \\ thus \: then \: accordingly \\ \\ = \frac{sin {}^{2} \: A + cos {}^{2} \:A - 1 - 2.sin \: A.cos \: A }{sin \: {}^{2} \:A - cos {}^{2} \: A - 1 + 2.cos \: A} \\ \\ we \: have \\ sin {}^{2} \: A + cos {}^{2} \: A = 1 \\ \\ thus \: then \\ \\ = \frac{1 - 1 - 2.sin \:A.cos \: A}{1 - cos {}^{2} \: A - cos {}^{2} \:A - 1 + 2.cos \:A } \\ \\ = \frac{ - 2.sin \: A.cos \:A }{ - 2.cos {}^{2} \:A + 2.cos \: A } \\ \\ \\ = \frac{ - 2.sin \:A.cos \: A }{ - 2.cos \: A(cos \:A - 1) } \\ \\ \\ = \frac{sin \: A}{cos \: A - 1} \\ \\ \\ = \frac{sin \: A}{cos \:A - 1 } \times \frac{cos \:A + 1 }{cos \: A + 1} \\ \\ = \frac{sin \: A(cos \:A + 1) }{cos {}^{2} \: A - 1} \\ \\ \\ = \frac{sin \: A(1 + cos \:A) }{ - sin {}^{2} \: A } \\ \\ = \frac{1 + cos \: A}{( - sin \:A) }$

$\\ = \frac{1}{( - sin \: A)} + \frac{cos \: A}{ (- sin \:A) } \\ \\ = - cosec \: A - cot \: A \\ \\ = - (cosec \: A + cot \: A)$