Let S={theta in[0,2 pi): tan (pi cos theta)+t | vedclass

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(D) Given $	an (\pi \cos 	heta) + 	an (\pi \sin 	heta) = 0$.This implies $	an (\pi \cos 	heta) = -	an (\pi \sin 	heta) = 	an (-\pi \sin 	het

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$2$

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(D) Given $	an (\pi \cos 	heta) + 	an (\pi \sin 	heta) = 0$.This implies $	an (\pi \cos 	heta) = -	an (\pi \sin 	heta) = 	an (-\pi \sin 	heta)$.Thus,$\pi \cos 	heta = n\pi - \pi \sin 	heta$,where $n \in \mathbb{Z}$.Dividing by $\pi$,we get $\sin 	heta + \cos 	heta = n$.Since $-\sqrt{2} \leq \sin 	heta + \cos 	heta \leq \sqrt{2}$,the possible integer values for $n$ are $-1, 0, 1$.Case $1$: $\sin 	heta + \cos 	heta = 0 \implies 	an 	heta = -1$. In $[0, 2\pi)$,$	heta = \frac{3\pi}{4}, \frac{7\pi}{4}$.Case $2$: $\sin 	heta + \cos 	heta = 1 \implies \sin(	heta + \frac{\pi}{4}) = \frac{1}{\sqrt{2}}$. In $[0, 2\pi)$,$	heta = 0, \frac{\pi}{2}$.Case $3$: $\sin 	heta + \cos 	heta = -1 \implies \sin(	heta + \frac{\pi}{4}) = -\frac{1}{\sqrt{2}}$. In $[0, 2\pi)$,$	heta = \pi, \frac{3\pi}{2}$.The set $S = \{0, \frac{\pi}{2}, \frac{3\pi}{4}, \pi, \frac{3\pi}{2}, \frac{7\pi}{4}\}$.We need to calculate $\sum_{	heta \in S} \sin^2(	heta + \frac{\pi}{4})$.For $	heta \in \{0, \frac{\pi}{2}\}$,$\sin^2(	heta + \frac{\pi}{4}) = (\frac{1}{\sqrt{2}})^2 = \frac{1}{2}$.For $	heta \in \{\pi, \frac{3\pi}{2}\}$,$\sin^2(	heta + \frac{\pi}{4}) = (-\frac{1}{\sqrt{2}})^2 = \frac{1}{2}$.For $	heta \in \{\frac{3\pi}{4}, \frac{7\pi}{4}\}$,$\sin^2(	heta + \frac{\pi}{4}) = (0)^2 = 0$.Sum $= 4 	imes \frac{1}{2} + 2 	imes 0 = 2$.

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

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