Question

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

Answer 1

LHS = `(sin theta)/((sec theta + tan theta -1)) + cos theta/((cosec theta + cot theta -1))`

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

      =`sin theta cos theta [1/(1+ (sin theta - cos theta))+ 1/(1- (sin theta - cos theta))]`

      =`sin theta cos theta [(1-(sin theta - cos theta)+1+(sin theta - cos theta))/({1+ (sin theta - cos theta )}{1- (sin theta-cos theta)})]`

     =`sin theta cos theta[(1-sin theta + cos theta +1+sin theta - cos theta)/(1-(sin theta - cos theta)^2)]`

     =`(2 sin theta cos theta)/(1-(sin ^2 theta + cos^2 theta -2 sin theta cos theta))`

    =`(2 sin theta cos theta )/(2 sin theta cos theta)`

    =1

    = RHS
Hence, LHS = RHS