If 1 + sin x + sin^2 x + dots infty = 4 + 2sq | vedclass

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(D) The given expression is an infinite geometric series with first term $a = 1$ and common ratio $r = \sin x$.For the sum to exist,$|\sin x| < 1$.The

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none of these

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(D) The given expression is an infinite geometric series with first term $a = 1$ and common ratio $r = \sin x$.For the sum to exist,$|\sin x| The sum of an infinite geometric series is $S = \frac{a}{1 - r}$.Thus,$\frac{1}{1 - \sin x} = 4 + 2\sqrt{3}$.$1 - \sin x = \frac{1}{4 + 2\sqrt{3}}$.Rationalizing the denominator: $\frac{1}{4 + 2\sqrt{3}} 	imes \frac{4 - 2\sqrt{3}}{4 - 2\sqrt{3}} = \frac{4 - 2\sqrt{3}}{16 - 12} = \frac{4 - 2\sqrt{3}}{4} = 1 - \frac{\sqrt{3}}{2}$.So,$1 - \sin x = 1 - \frac{\sqrt{3}}{2}$.$\sin x = \frac{\sqrt{3}}{2}$.In the interval $(0, \pi)$,$\sin x = \frac{\sqrt{3}}{2}$ at $x = \frac{\pi}{3}$ and $x = \frac{2\pi}{3}$.Since neither $\frac{\pi}{3}$ nor $\frac{2\pi}{3}$ matches the given options $A, B, C$,the correct answer is $D$.

If $1 + \sin x + \sin^2 x + \dots \infty = 4 + 2\sqrt{3}$,where $0 < x < \pi$,then:

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