Question

Prove that : `(sectheta+tantheta-1)/(tantheta-sectheta+1)=(costheta)/(1-sintheta)`

Answer 1

LHS `= (sec theta + tan theta -1)/(tan theta - sec theta +1) `
` = ((sec theta + tan theta)-(sec^(2)theta - tan^(2)theta))/((tan theta - sec theta +1)) " " [because sec^(2)theta - tan^(2)theta =1] `
`=((sec theta + tan theta)[1-(sec theta - tan theta)])/((tan theta - sec theta +1))`
`= ((sec theta + tan theta )(tan theta - sec theta +1))/((tan theta - sec theta +1))=(sec theta + tan theta)`
` =((1)/(cos theta)+(sin theta)/(cos theta))=((1+ sin theta)/(cos theta))=((1+ sin theta))/(cos theta) xx ((1- sin theta))/((1- sin theta)) `
` = ((1- sin^(2)theta))/(cos theta(1- sin theta)) = (cos^(2)theta)/(cos theta(1- sin theta))=(cos theta)/((1-sin theta))= RHS. `
Hence, ` (sec theta + tan theta -1)/(tan theta - sec theta +1)=(cos theta)/((1- sin theta)). `