Question

`cos^2 theta /((1 tan theta))+ sin ^3 theta/((sin theta - cos theta))=(1+sin theta cos theta)`

Answer 1

`cos^2 theta /((1 tan theta))+ sin ^3 theta/((sin theta - cos theta))=(1+sin theta cos theta)`

LHS=`cos^2theta/((1-tan theta))+sin ^3theta/((sin theta - cos theta))`

     =`cos^2theta/(1-sintheta/costheta)+sin^3 theta/((sin theta-costheta))`

     =`cos^3 theta/((cos theta-sin theta))+ sin ^3 theta/((sintheta-cos theta))`

     =`(cos^3theta-sin^3 theta)/((costheta - sin theta))`

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

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

  =`(1+sin theta cos theta)`

   =RHS

Hence, L.H.S = R.H.S.