Given that,
$$
\begin{array}{cc}
& \tan \theta+\sin \theta=x \\
& \tan \theta-\sin \theta=y \\
& \tan \theta=\frac{x+y}{2}, \sin \theta=\frac{x-y}{2} \\
\therefore \quad & \left(\frac{x+y}{2}\right)\left(\frac{x-y}{2}\right)=\tan \theta \sin \theta \\
\Rightarrow \quad & x^2-y^2=4\left(\frac{\sin ^2 \theta}{\cos \theta}\right)=4\left(\frac{1-\cos ^2 \theta}{\cos \theta}\right) \\
\Rightarrow \quad & x^2-y^2=4(\sec \theta-\cos \theta) \\
\Rightarrow \quad & \left(x^2-y^2\right)^2=16\left(\sec \theta^2+\cos ^2 \theta-2\right) \\
\Rightarrow \quad & \left(x^2-y^2\right)^2=16\left(1+\tan ^2 \theta+1-\sin ^2 \theta-2\right) \\
\Rightarrow \quad & \left(x^2-y^2\right)^2=16\left(\tan { }^2 \theta-\sin ^2 \theta\right) \\
& \quad\left(x^2-y^2\right)^2=16 x y
\end{array}
$$