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(D) The given series is an infinite geometric progression with first term $a = 1$ and common ratio $r = \sin 	heta$.Since the sum to infinity is $S =

Vedclass · Accepted Answer

$\frac{\pi}{3} 	ext{ or } \frac{2\pi}{3}$

Vedclass · Answer

(D) The given series is an infinite geometric progression with first term $a = 1$ and common ratio $r = \sin 	heta$.Since the sum to infinity is $S = \frac{a}{1-r}$,we have $\frac{1}{1 - \sin 	heta} = 4 + 2\sqrt{3}$.Taking the reciprocal,$1 - \sin 	heta = \frac{1}{4 + 2\sqrt{3}}$.Rationalizing the denominator: $1 - \sin 	heta = \frac{4 - 2\sqrt{3}}{(4 + 2\sqrt{3})(4 - 2\sqrt{3})} = \frac{4 - 2\sqrt{3}}{16 - 12} = \frac{4 - 2\sqrt{3}}{4} = 1 - \frac{\sqrt{3}}{2}$.Thus,$\sin 	heta = \frac{\sqrt{3}}{2}$.Given $0 < 	heta < \pi$ and $	heta 
eq \frac{\pi}{2}$,the possible values for $	heta$ are $\frac{\pi}{3}$ and $\pi - \frac{\pi}{3} = \frac{2\pi}{3}$.

If $1 + \sin \theta + \sin^2 \theta + \dots \text{ to } \infty = 4 + 2\sqrt{3}$,where $0 < \theta < \pi$ and $\theta \neq \frac{\pi}{2}$,then $\theta = $

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