Obviously,
\(E_{p\,e}=h.\left\{1-\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}\right\}.f_{cir}\)
\(E_{p\,a}=h.\left\{1-\sqrt{\cfrac{cos(\theta)}{sin(\theta)}}\right\}.\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}f_{cir}=h.\left\{\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}-1\right\}f_{cir}=-E_{p\,e}\)
The emitted photon is exactly the absorbed photon.
