If tan theta = sqrt{frac{3}{2}},the sum of th | vedclass

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(D) Let the given series be $S = 1 + 2x + 3x^2 + 4x^3 + \dots \infty$,where $x = (1 - \cos 	heta)$.This is an arithmetico-geometric series of the for

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$\frac{5}{2}$

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(D) Let the given series be $S = 1 + 2x + 3x^2 + 4x^3 + \dots \infty$,where $x = (1 - \cos 	heta)$.This is an arithmetico-geometric series of the form $\sum_{n=1}^{\infty} n x^{n-1}$.The sum of this infinite series is given by $S = (1 - x)^{-2}$ for $|x| Substituting $x = 1 - \cos 	heta$ into the formula:$S = (1 - (1 - \cos 	heta))^{-2} = (\cos 	heta)^{-2} = \sec^2 	heta$.Using the trigonometric identity $\sec^2 	heta = 1 + 	an^2 	heta$:$S = 1 + 	an^2 	heta$.Given $	an 	heta = \sqrt{\frac{3}{2}}$,then $	an^2 	heta = \frac{3}{2}$.Therefore,$S = 1 + \frac{3}{2} = \frac{5}{2}$.

If $\tan \theta = \sqrt{\frac{3}{2}}$,the sum of the infinite series $1 + 2(1 - \cos \theta) + 3(1 - \cos \theta)^2 + 4(1 - \cos \theta)^3 + \dots \infty$ is

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