If theta = frac{pi}{6},then the 10^{text{th}} | vedclass

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(B) The given series is a geometric progression with the first term $a = 1$ and common ratio $r = (\cos 	heta + i \sin 	heta)$.The $n^{	ext{th}}$ t

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$-i$

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(B) The given series is a geometric progression with the first term $a = 1$ and common ratio $r = (\cos 	heta + i \sin 	heta)$.The $n^{	ext{th}}$ term of a geometric progression is given by $T_n = a r^{n-1}$.For the $10^{	ext{th}}$ term $(n = 10)$:$T_{10} = 1 	imes (\cos 	heta + i \sin 	heta)^{10-1} = (\cos 	heta + i \sin 	heta)^9$.Using De Moivre's Theorem,$(\cos 	heta + i \sin 	heta)^n = \cos(n	heta) + i \sin(n	heta)$.$T_{10} = \cos(9	heta) + i \sin(9	heta)$.Given $	heta = \frac{\pi}{6}$,we have $9	heta = 9 	imes \frac{\pi}{6} = \frac{3\pi}{2}$.$T_{10} = \cos(\frac{3\pi}{2}) + i \sin(\frac{3\pi}{2})$.Since $\cos(\frac{3\pi}{2}) = 0$ and $\sin(\frac{3\pi}{2}) = -1$,$T_{10} = 0 + i(-1) = -i$.

If $\theta = \frac{\pi}{6}$,then the $10^{\text{th}}$ term of the series $1 + (\cos \theta + i \sin \theta)^1 + (\cos \theta + i \sin \theta)^2 + \dots$ is

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