If n is an integer and Z = cos theta + i sin | vedclass

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(D) Given $Z = \cos 	heta + i \sin 	heta$. By De Moivre's Theorem,$Z^{2n} = \cos(2n	heta) + i \sin(2n	heta)$. Substituting this into the expressio

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$-i \cot n 	heta$

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(D) Given $Z = \cos 	heta + i \sin 	heta$. By De Moivre's Theorem,$Z^{2n} = \cos(2n	heta) + i \sin(2n	heta)$. Substituting this into the expression: $\frac{1 + Z^{2n}}{1 - Z^{2n}} = \frac{1 + \cos(2n	heta) + i \sin(2n	heta)}{1 - (\cos(2n	heta) + i \sin(2n	heta))} = \frac{1 + \cos(2n	heta) + i \sin(2n	heta)}{1 - \cos(2n	heta) - i \sin(2n	heta)}$. Using half-angle identities $1 + \cos(2A) = 2 \cos^2 A$ and $1 - \cos(2A) = 2 \sin^2 A$,and $\sin(2A) = 2 \sin A \cos A$: $= \frac{2 \cos^2(n	heta) + 2i \sin(n	heta) \cos(n	heta)}{2 \sin^2(n	heta) - 2i \sin(n	heta) \cos(n	heta)} = \frac{2 \cos(n	heta) [\cos(n	heta) + i \sin(n	heta)]}{2i \sin(n	heta) [-i \sin(n	heta) + \cos(n	heta)]}$. $= \frac{\cos(n	heta)}{i \sin(n	heta)} = -i \cot(n	heta)$.

If $n$ is an integer and $Z = \cos \theta + i \sin \theta$,where $\theta \neq (2n + 1) \frac{\pi}{2}$,then $\frac{1 + Z^{2n}}{1 - Z^{2n}} = $

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