Question

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

Answer 1

We have , `(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^2 theta - cos ^2 theta)`

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

     =`2/(sin^2 theta - cos^2 theta)`

Again ,` 2/(sin^2 theta - cos^2 theta)`

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

   =`2/(2 sin ^2 theta -1)`