Let $S=\left\{\theta \in [0,2\mathrm{\pi }): \tan \left(\mathrm{\pi cos\theta }\right)+\tan \left(\mathrm{\pi sin\theta }\right)=0\right\}$, then $\sum _{\theta \in S}{\sin }^2\left(\theta +\frac{\mathrm{\pi }}{4}\right)$ is equal to

Given:

$S=\left\{\theta \in [0,2\mathrm{\pi }): \tan \left(\mathrm{\pi cos\theta }\right)+\tan \left(\mathrm{\pi sin\theta }\right)=0\right\}$

So,

$\tan \left(\mathrm{\pi cos\theta }\right)+\tan \left(\mathrm{\pi sin\theta }\right)=0$

$\Rightarrow \tan \left(\mathrm{\pi cos\theta }\right)=-\tan \left(\mathrm{\pi sin\theta }\right)$

$\Rightarrow \tan \left(\mathrm{\pi cos\theta }\right)=\tan \left(-\mathrm{\pi sin\theta }\right)$

$\Rightarrow \mathrm{\pi cos\theta }=n\mathrm{\pi }-\mathrm{\pi sin\theta }; n\in Z$

$\Rightarrow \mathrm{sin\theta }+\mathrm{cos\theta }=n$

Now,

$-\sqrt{2}\le \mathrm{sin\theta }+\mathrm{cos\theta }\le \sqrt{2}$

$\Rightarrow -\sqrt{2}\le n\le \sqrt{2}$

But $n\in Z$, so $n=-1, 0, 1$

So,

$\theta \in \left\{0,\frac{\mathrm{\pi }}{2},\frac{3\mathrm{\pi }}{4},\frac{7\mathrm{\pi }}{4},\frac{3\mathrm{\pi }}{2},\mathrm{\pi }\right\}$

So,

$\sum _{\theta \in S}{\sin }^2\left(\theta +\frac{\mathrm{\pi }}{4}\right)=\frac{1}{2}+\frac{1}{2}+0+0+\frac{1}{2}+\frac{1}{2}=2$