If$x=\tan \theta +\sin \theta ,y=\tan \theta -\sin \theta$ then ${\left(x^2-y^2\right)}^2=$

$$\begin{gathered} x+y=2\tan \theta , \\ x-y=2\sin \theta , \\ we know that \\ {\tan }^2\theta -{\sin }^2\theta ={\tan }^2\theta {\sin }^2\theta \end{gathered}$$ 

$$\begin{gathered} \Rightarrow {\left(\frac{x+y}{2}\right)}^2-{\left(\frac{x-y}{2}\right)}^2={\left(\frac{x+y}{2}\right)}^2.{\left(\frac{x+y}{2}\right)}^2 \\ \Rightarrow xy=\frac{{\left(x^2-y^2\right)}^2}{16} \end{gathered}$$

${\left(x^2-y^2\right)}^2=16xy.$