Question

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

Answer 1

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

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

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

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

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

    =` 2/ (sin^2 theta -1)`

    = RHS