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

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Question

$	an (\pi \cos 	heta)+	an (\pi \sin 	heta)=0$

$	an (\pi \cos 	heta)=-	an (\pi \sin 	heta)$

$	an (\pi \cos 	heta)=	an (-\pi \sin 	het

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$	an (\pi \cos 	heta)+	an (\pi \sin 	heta)=0$

$	an (\pi \cos 	heta)=-	an (\pi \sin 	heta)$

$	an (\pi \cos 	heta)=	an (-\pi \sin 	heta)$

$\pi \cos 	heta=n \pi-\pi \sin 	heta$

$\sin 	heta+\cos 	heta= n 	ext { where } n \in I$

possible values are $n =0,1$ and $-1$ because

$-\sqrt{2} \leq \sin 	heta+\cos 	heta \leq \sqrt{2}$

Now it gives $	heta \in\left\{0, \frac{\pi}{2}, \frac{3 \pi}{4}, \frac{7 \pi}{4}, \frac{3 \pi}{2}, \piight\}$

So $\sum_{	heta \in S} \sin ^2\left(	heta+\frac{\pi}{4}ight)=2(0)+4\left(\frac{1}{2}ight)=2$

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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