Let S = left{ theta in [-pi, pi] - left{ pm f | vedclass

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(B) Given equation: $\sin 	heta 	an 	heta + 	an 	heta = \sin 2 	heta$$	an 	heta (\sin 	heta + 1) = \frac{2 	an 	heta}{1 + 	an^2 	heta} =

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$9$

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(B) Given equation: $\sin 	heta 	an 	heta + 	an 	heta = \sin 2 	heta$$	an 	heta (\sin 	heta + 1) = \frac{2 	an 	heta}{1 + 	an^2 	heta} = 2 \sin 	heta \cos 	heta$Case $1$: $	an 	heta = 0 \implies 	heta = 0, \pi, -\pi$ (since $	heta \in [-\pi, \pi]$).Case $2$: $\sin 	heta + 1 = 2 \cos^2 	heta = 2(1 - \sin^2 	heta) = 2(1 - \sin 	heta)(1 + \sin 	heta)$.If $\sin 	heta = -1$,then $	heta = -\frac{\pi}{2}$,which is excluded.If $\sin 	heta 
eq -1$,then $1 = 2(1 - \sin 	heta) \implies 1 = 2 - 2 \sin 	heta \implies \sin 	heta = \frac{1}{2}$.Thus,$	heta = \frac{\pi}{6}, \frac{5\pi}{6}$.The set $S = \{0, \pi, -\pi, \frac{\pi}{6}, \frac{5\pi}{6} \}$,so $n(S) = 5$.$T = \sum_{	heta \in S} \cos 2 	heta = \cos(0) + \cos(2\pi) + \cos(-2\pi) + \cos(\frac{\pi}{3}) + \cos(\frac{5\pi}{3})$$T = 1 + 1 + 1 + \frac{1}{2} + \frac{1}{2} = 4$.Therefore,$T + n(S) = 4 + 5 = 9$.

Let $S = \left\{ \theta \in [-\pi, \pi] - \left\{ \pm \frac{\pi}{2} \right\} : \sin \theta \tan \theta + \tan \theta = \sin 2 \theta \right\}$. If $T = \sum_{\theta \in S} \cos 2 \theta$,then $T + n(S)$ is equal to:

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