The equation of the tangent to the curve $x(\theta)=2 \sqrt{2} (\cos \theta+\theta \sin \theta), y(\theta)=2 \sqrt{2}(\sin \theta-\theta \cos \theta)$, at $\theta=\frac{\pi}{4}$ is equal to

$x-y=\pi$

$x(\theta)=2 \sqrt{2} (\cos \theta+\theta \sin \theta), ~~y(\theta)=2 \sqrt{2}(\sin \theta-\theta \cos \theta)$

$x_0=x(\pi / 4)=2 \sqrt{2}\left[\frac{1}{\sqrt{2}}+\frac{\pi}{4 ~\sqrt{2}}\right] ,  ~~y(\pi / 4)=2 \sqrt{2}\left[\frac{1}{\sqrt{2}}-\frac{\pi}{4 ~\sqrt{2}}\right]=y_0$

$\frac{d y}{d x}=\frac{2 \sqrt{2}(\cos \theta-\cos \theta+\theta \sin \theta)}{2 \sqrt{2}(-\sin \theta+\sin \theta+\theta \cos \theta)} = \tan \theta \Rightarrow\left.\frac{d y}{d x}\right]_{\theta=\pi / 4}=1$   →   slope

so  $y-y_0 = 1 × \left(x-x_0\right)$

so  $x-y =x_0-y_0$

$x-y =\pi$

Option: B