Let S = \ θ ∈ [ 0, 2π ) : (π cos θ) + (π sin θ) = 0\ , then ∑ _ θ ∈...

Given:

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

So,

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

$⇒ \tan \left(\pi cos \theta\right) = - \tan \left(\pi sin \theta\right)$

$⇒ \tan \left(\pi cos \theta\right) = \tan \left(- \pi sin \theta\right)$

$⇒ \pi cos \theta = n\pi - \pi sin \theta; n \in Z$

$⇒ sin \theta + cos \theta = n$

Now,

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

$⇒ - \sqrt{2} \le n \le \sqrt{2}$

But $n \in Z$, so $n = - 1, 0, 1$

So,

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

So,

$\mathop{∑}_{\theta \in S}\sin^{2}\left(\theta + \dfrac{\pi}{4}\right) = \dfrac{1}{2} + \dfrac{1}{2} + 0 + 0 + \dfrac{1}{2} + \dfrac{1}{2} = 2$