Prove





[tex]\small \dfrac{ \sin(A) - \cos(A) + 1}{\sin(A) + \cos(A) - 1} = \dfrac{\cos(A)}{1 - \sin(A) } [/tex]




Help!!!!!!!!!​

Respuesta :

[tex] \frac{ \sin(a) - \cos(a) + 1 }{ \sin(a) + \cos(a) - 1 } = \\ [/tex]

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[tex] \frac{ \sin(a) - \cos(a) + 1 }{ \sin(a) + \cos(a) - 1 } \times \frac{ \sin(a) + \cos(a) + 1}{ \sin(a) + \cos(a) + 1 } = [/tex]

[tex] \frac{ {sin}^{2}(a) + 2 \sin(a) - {cos}^{2} (a) + 1 }{ {sin}^{2}(a) + 2 \sin(a) \cos(a) + {cos}^{2}(a) - 1 } = [/tex]

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As you know :

[tex] {sin}^{2} (a) + {cos}^{2} (a) = 1[/tex]

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[tex] \frac{ {sin}^{2} (a) - {cos}^{2}(a) + 2 \sin(a) + 1}{ {sin}^{2} (a) + {cos}^{2}(a) - 1 + 2 \sin(a) \cos(a) } = [/tex]

[tex] \frac{ {sin}^{2}(a) - (1 - {sin}^{2}(a)) + 2 \sin(a) + 1 }{1 - 1 + 2 \sin(a) \cos(a) } = [/tex]

[tex] \frac{ {sin}^{2} (a) + {sin}^{2} (a) - 1 + 1 + 2 \sin(a) }{2 \sin(a) \cos(a) } = [/tex]

[tex] \frac{2 {sin}^{2}(a) + 2 \sin(a) }{2 \sin(a) \cos(a) } = [/tex]

[tex] \frac{2 \sin(a)( \sin(a) + 1) }{2 \sin(a)( \cos(a) \: ) } = \\ [/tex]

[tex] \frac{ \sin(a) + 1}{ \cos(a) } \\ [/tex]

And we're done...

Take care ♡♡♡♡♡