Question

prove that : sina-cosa+1/sina+cosa-1 = 1/seca-tana​

Answer 1

LHS = sinA-cosA+1/sinA+cosA-1

divide both numerator and denominator by cosA

LHS=(tanA−1+secA)/(tanA+1−secA)

Now

sec2A=1+tan2A

sec2A−tan2A=1

Using above relation at denominator of LHS

LHS=(tanA−1+secA)/(tanA−secA+sec2A−tan2A)

LHS=(tanA−1+secA)/((secA−tanA)(−1+secA+tanA))

LHS=1/(secA−tanA)

LHS=RHS

Hence Proved.

Answer 2

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

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