If left(frac{cos theta+i sin theta}{sin theta | vedclass

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(C) Given $	heta = \frac{\pi}{2}$,we have $\cos \frac{\pi}{2} = 0$ and $\sin \frac{\pi}{2} = 1$. Substituting these values into the first term: $\lef

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$2$

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(C) Given $	heta = \frac{\pi}{2}$,we have $\cos \frac{\pi}{2} = 0$ and $\sin \frac{\pi}{2} = 1$. Substituting these values into the first term: $\left(\frac{0+i(1)}{1+i(0)}ight)^{2024} = (i)^{2024} = (i^4)^{506} = 1^{506} = 1$. For the second term: $\frac{1+\cos 	heta+i \sin 	heta}{1-\cos 	heta+i \sin 	heta} = \frac{1+0+i(1)}{1-0+i(1)} = \frac{1+i}{1+i} = 1$. Thus,the expression becomes $1^{2024} + 1^{2025} = 1 + 1 = 2$. Since $x+iy = 2$,we have $x=2$ and $y=0$. Therefore,$x+y = 2+0 = 2$.

If $\left(\frac{\cos \theta+i \sin \theta}{\sin \theta+i \cos \theta}\right)^{2024}+\left(\frac{1+\cos \theta+i \sin \theta}{1-\cos \theta+i \sin \theta}\right)^{2025}=x+i y$,then the value of $x+y$ at $\theta=\frac{\pi}{2}$ is

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