If $\tan (\pi \cos \theta)=\cot (\pi \sin \theta)$, then $\sin \left(\frac{\pi}{4}+\theta\right)=$

The equation $\tan (\pi \cos \theta) = \cot (\pi \sin \theta)$ is solved by rewriting $\cot x = \tan \left(\frac{\pi}{2} - x\right)$, giving $\tan (\pi \cos \theta) = \tan \left(\frac{\pi}{2} - \pi \sin \theta\right)$.

Applying the general solution for $\tan A = \tan B$ as $A = n\pi + B$ for integer $n$, we have $\pi \cos \theta = n\pi + \left(\frac{\pi}{2} - \pi \sin \theta\right)$. Dividing by $\pi$ and rearranging yields $\cos \theta + \sin \theta = n + \frac{1}{2}$.

Since the range of $\cos \theta + \sin \theta$ is $[-\sqrt{2}, \sqrt{2}]$, it follows that $-\sqrt{2} \leq n + \frac{1}{2} \leq \sqrt{2}$. Solving the inequality gives $-1.914 \leq n \leq 0.914$, so the integer solutions are $n = -1$ and $n = 0$.

This implies $\cos \theta + \sin \theta = -\frac{1}{2}$ for $n = -1$, and $\frac{1}{2}$ for $n = 0$.

Using the identity $\sin \left(\frac{\pi}{4} + \theta\right) = \frac{1}{\sqrt{2}} (\cos \theta + \sin \theta)$, we find $\sin \left(\frac{\pi}{4} + \theta\right) = -\frac{1}{2\sqrt{2}}$ or $\frac{1}{2\sqrt{2}}$.

The value $\frac{1}{2\sqrt{2}}$ corresponds to option D.