A) \[\theta =\frac{\pi }{3}\]
B) \[\theta =\frac{\pi }{6}\]
C) \[\theta =\frac{\pi }{3}\]or\[\frac{\pi }{6}\]
D) \[\theta =\frac{\pi }{3}\]or\[\frac{2\pi }{3}\]
Correct Answer: D
Solution :
Given, \[1+\sin \theta +{{\sin }^{2}}\theta +...+\infty =4+2\sqrt{3}\] \[\Rightarrow \] \[\frac{1}{1-\sin \theta }=4+2\sqrt{3}\,\,(\because 0<\sin \theta <1)\] \[\Rightarrow \] \[1-\sin \theta =\frac{1}{4-2\sqrt{3}}\times \frac{4-2\sqrt{3}}{4-2\sqrt{3}}\] \[\Rightarrow \] \[1-\sin \theta =\frac{4-2\sqrt{3}}{16-12}=1-\frac{\sqrt{3}}{2}\] \[\Rightarrow \] \[\sin \theta =\frac{\sqrt{3}}{2}=\sin \frac{\pi }{3}\] \[\theta =\frac{\pi }{3}\] or \[\theta =\pi -\frac{\pi }{3}\] \[=\frac{2\pi }{3}\]
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