[(sin theta-cos theta+1)/(sin theta+cos theta-1)=],[(1)/(sec theta-tan theta)],[sec^(2)theta=1+tan^(2)theta)]

(sin theta-cos theta + 1) / (sin theta + cos theta-1) = sec theta + tan theta

((sin theta+cos theta)^(2)-1)/(sin theta*cos theta)

(sin theta-cos theta+1)/(sin theta+cos theta-1)=(1+sin theta)/(cos theta)

Prove that quad (sin theta-cos theta+1)/(sin theta+cos theta-1)=(1)/(sec theta-tan theta) ,using the identity sec^(2)theta=1+tan^(2)theta

((sin theta+cos theta-1)/(sin theta-cos theta+1))(1+sin theta) =

(1 + sin theta-cos theta) / (1 + sin theta + cos theta)

Prove that: (sin theta - cos theta +1)/(sin theta + cos theta -1) = (1)/(sec theta - tan theta)

Prove that (sin theta-cos theta+1)/(sin theta+cos theta-1)=1/(sec theta-tan theta) ,using the identity sec^2 theta=1+tan^2 theta .

(1 + sin theta-cos theta) / (1 + sin theta + cos theta) =

Prove that (sin theta - cos theta +1)/(sin theta + cos theta -1) = (1)/((sec theta - tan theta)).