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

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Question

$\sin 	heta 	an 	heta+	an 	heta=\sin 2 	heta$

$	an 	heta(\sin 	heta+1)=\frac{2 	an 	heta}{1+	an ^{2} 	heta}$

$	an 	heta=0 \Righta

Vedclass · Accepted Answer

$9$

Vedclass · Answer

$\sin 	heta 	an 	heta+	an 	heta=\sin 2 	heta$

$	an 	heta(\sin 	heta+1)=\frac{2 	an 	heta}{1+	an ^{2} 	heta}$

$	an 	heta=0 \Rightarrow 	heta=-\pi, 0, \pi$

$(\sin 	heta+1)=2 \cdot \cos ^{2} 	heta=2(1+\sin 	heta)(1-\sin 	heta)$

$\sin 	heta=-1$ which is not possible

$\sin 	heta=\frac{1}{2} \quad 	heta=\frac{\pi}{6}, \frac{5 \pi}{6}$

$n ( s )=5$

$T =\cos 0+\cos 2 \pi+\cos 2 \pi+\cos \frac{\pi}{3}+\cos \frac{5 \pi}{3}$

$T =4$

$T + n ( s )=9$

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

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