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જો $1\,\, + \,\,\sin \theta \,\, + \,\,{\sin ^2}\theta + \ldots .\,\,to\,\,\infty \,\, = \,\,4\, + 2\sqrt 3 ,\,\,0\,\, < \,\theta \,\,\pi ,\,\,\theta \,\, \ne \,\frac{\pi }{2}\,,$ હોય તો $\theta = $
$\frac{\pi }{6}$
$\frac{\pi }{3}$
$\frac{\pi }{3}$ or $\frac{\pi }{6}$
$\frac{\pi }{3}$ or $\frac{2\pi }{3}$
Solution
${1+\sin \theta+\sin ^{2} \theta+\ldots \ldots=4+2 \sqrt{3}} $
${\Rightarrow \quad \frac{1}{1-\sin \theta}=4+2 \sqrt{3}} $
${\Rightarrow \quad \frac{1}{1-\sin \theta}=\frac{1}{4+2 \sqrt{3}}} $
${\Rightarrow \quad 1-\sin \theta=\frac{1}{2(2+\sqrt{3})(2-\sqrt{3})}} $
${=\frac{1}{2} \cdot \frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}} $
${=\frac{1}{2} \cdot \frac{2-\sqrt{3}}{4-3}=\frac{2-\sqrt{3}}{2}=1-\frac{\sqrt{3}}{2}}$
${\Rightarrow \quad \sin \theta=\frac{\sqrt{3}}{2} \quad \Rightarrow \theta=\frac{\pi}{6}}$
