Question

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

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

$\frac{ \sin(A) - \cos(A) + 1 }{ \sin(A) + \cos(A) - 1} \\ \\ = \frac{ \frac{ \sin(A) - \cos(A) + 1 }{ \cos(A) } }{ \frac{ \sin(A) + \cos(A) - 1 }{ \cos(A) } } \\ \\ = \frac{ \tan(A) + \sec(A) - 1 }{ \tan(A) - \sec(A) + 1 }$

$= \frac{ \tan(A) + \sec(A) - 1}{( { \sec }^{2}A - { \tan}^{2}A) - ( \sec A - \tan A)} \\ \\ = \frac{ \tan A + \sec A - 1}{( \sec A - \tan A)( \sec A + \tan A - 1)} \\ \\ = \frac{1}{ \sec(A) - \tan(A) }$