Question

Prove that
SinA - cosA + 1 / sinA + cosA - 1 = 1 / secA - tanA

Answer 1

Answer:

LHS = (sinA-cosA+1)/(sinA+cos-1)

        = (1 + sinA) (1 - sinA)/cos A(1 - sinA)

        = 1 - sin2A/cos A(1 - sinA)

        = cos2A/cos A(1 - sinA)

        = cosA/(1 - sinA)

        = 1/ (1/cosA - sinA/cosA)

        = 1/(secA - tanA) = (RHS Hence Proved)

Answer 2

$\sf{\blue{See \: Attachment}}$