Question

Prove

Answer 1

$\frac{ \sin(a) - \cos(a) + 1 }{ \sin(a) + \cos(a) - 1 } =$

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$\frac{ \sin(a) - \cos(a) + 1 }{ \sin(a) + \cos(a) - 1 } \times \frac{ \sin(a) + \cos(a) + 1}{ \sin(a) + \cos(a) + 1 } =$

$\frac{ {sin}^{2}(a) + 2 \sin(a) - {cos}^{2} (a) + 1 }{ {sin}^{2}(a) + 2 \sin(a) \cos(a) + {cos}^{2}(a) - 1 } =$

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As you know :

${sin}^{2} (a) + {cos}^{2} (a) = 1$

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$\frac{ {sin}^{2} (a) - {cos}^{2}(a) + 2 \sin(a) + 1}{ {sin}^{2} (a) + {cos}^{2}(a) - 1 + 2 \sin(a) \cos(a) } =$

$\frac{ {sin}^{2}(a) - (1 - {sin}^{2}(a)) + 2 \sin(a) + 1 }{1 - 1 + 2 \sin(a) \cos(a) } =$

$\frac{ {sin}^{2} (a) + {sin}^{2} (a) - 1 + 1 + 2 \sin(a) }{2 \sin(a) \cos(a) } =$

$\frac{2 {sin}^{2}(a) + 2 \sin(a) }{2 \sin(a) \cos(a) } =$

$\frac{2 \sin(a)( \sin(a) + 1) }{2 \sin(a)( \cos(a) \: ) } =$

$\frac{ \sin(a) + 1}{ \cos(a) }$

And we're done...

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