sin theta minus cos theta + 1 divided by sin theta + cos theta minus 1 equal to 1 by sec theta minus 10 theta
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Let theta =A.
(sinA-cosA+1)/(sinA+cosA-1)=1/(secA-tanA).
L.H.S.
=(sinA-cosA+1)/(sinA+cosA-1). Dividing in Nr and Dr by cosA.
=(tanA-1+secA)/(tanA+1-secA).
=(tanA+secA-1)/(tanA-secA+1). , putting 1= sec^2A-tan^2A=(secA-tanA).
(secA+tanA). in Dr.
=(tanA+secA-1)/{-(secA-tanA)+(secA-tanA).(secA+tanA)}.
=(tanA+secA-1)/(secA-tanA).(-1+secA+tanA).
= 1/(secA-tanA). Proved.
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