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If $\sin \theta_{1}+\sin \theta_{2}+\sin \theta_{3}=3,$ then $\cos \theta_{1}+\cos \theta_{2}$$+\cos \theta^{3}=$
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Since, $\sin \theta_{1}+\sin \theta_{2}+\sin \theta_{3}=3$
$\begin{array}{ll}\therefore & \sin \theta_{1}=\sin \theta_{2}=\sin \theta_{3}=1 \Rightarrow \theta_{1}=\theta_{2}=\theta_{3}= \\ & \frac{\pi}{2} \\ \therefore & \cos \theta_{1}+\cos \theta_{2}+\cos \theta_{3}=0\end{array}$
$\begin{array}{ll}\therefore & \sin \theta_{1}=\sin \theta_{2}=\sin \theta_{3}=1 \Rightarrow \theta_{1}=\theta_{2}=\theta_{3}= \\ & \frac{\pi}{2} \\ \therefore & \cos \theta_{1}+\cos \theta_{2}+\cos \theta_{3}=0\end{array}$
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