If 1+sin theta+sin ^{2} theta+ldots infty = 2 | vedclass

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(C) Given the infinite geometric series: $1+\sin 	heta+\sin ^{2} 	heta+\ldots \infty = 2 \sqrt{3}+4$.The sum of an infinite geometric series is give

Vedclass · Accepted Answer

$\frac{\pi}{3}$

Vedclass · Answer

(C) Given the infinite geometric series: $1+\sin 	heta+\sin ^{2} 	heta+\ldots \infty = 2 \sqrt{3}+4$.The sum of an infinite geometric series is given by $S_{\infty} = \frac{a}{1-r}$,where $a = 1$ and $r = \sin 	heta$.Thus,$\frac{1}{1-\sin 	heta} = 2 \sqrt{3}+4$.Taking the reciprocal: $1-\sin 	heta = \frac{1}{2 \sqrt{3}+4}$.Rationalizing the denominator: $1-\sin 	heta = \frac{2 \sqrt{3}-4}{(2 \sqrt{3}+4)(2 \sqrt{3}-4)} = \frac{2 \sqrt{3}-4}{12-16} = \frac{2 \sqrt{3}-4}{-4}$.Simplifying: $1-\sin 	heta = -\frac{\sqrt{3}}{2} + 1$.Therefore,$\sin 	heta = \frac{\sqrt{3}}{2}$.Since $\sin 	heta = \frac{\sqrt{3}}{2}$,the value of $	heta$ is $\frac{\pi}{3}$.

If $1+\sin \theta+\sin ^{2} \theta+\ldots \infty = 2 \sqrt{3}+4$,then $\theta = $

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