Question

Sin theta-cos they+1/sintheta+costheta-1=1/sec theta-tan they

Answer 1

Answer:

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.