$\begin{aligned} & \tan (\pi \cos \theta)+\tan (\pi \sin \theta)=0 \\\\ & \tan (\pi \cos \theta)=-\tan (\pi \sin \theta) \\\\ & \tan (\pi \cos \theta)=\tan (-\pi \sin \theta) \\\\ & \pi \cos \theta=\mathrm{n} \pi-\pi \sin \theta \\\\ & \sin \theta+\cos \theta=\mathrm{n} \text { where } \mathrm{n} \in \mathrm{I} \end{aligned}$ possible values are $\mathrm{n}=0,1$ and $-1$ because $-\sqrt{2} \leq \sin \theta+\cos \theta \leq \sqrt{2}$ Now it gives $\theta \in\left\{0, \frac{\pi}{2}, \frac{3 \pi}{4}, \frac{7 \pi}{4}, \frac{3 \pi}{2}, \pi\right\}$ So $\sum \limits_{\theta \in \mathrm{S}} \sin ^2\left(\theta+\frac{\pi}{4}\right)=2(0)+4\left(\frac{1}{2}\right)=2$