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(D) Given that $a = \cos \alpha + i\sin \alpha$,$b = \cos \beta + i\sin \beta$,and $c = \cos \gamma + i\sin \gamma$.Using Euler's formula,we have $a =

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$1$

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(D) Given that $a = \cos \alpha + i\sin \alpha$,$b = \cos \beta + i\sin \beta$,and $c = \cos \gamma + i\sin \gamma$.Using Euler's formula,we have $a = e^{i\alpha}$,$b = e^{i\beta}$,and $c = e^{i\gamma}$.Then,$\frac{b}{c} = e^{i(\beta - \gamma)} = \cos(\beta - \gamma) + i\sin(\beta - \gamma)$ $(i)$.Similarly,$\frac{c}{a} = e^{i(\gamma - \alpha)} = \cos(\gamma - \alpha) + i\sin(\gamma - \alpha)$ $(ii)$.And $\frac{a}{b} = e^{i(\alpha - \beta)} = \cos(\alpha - \beta) + i\sin(\alpha - \beta)$ $(iii)$.Adding $(i)$,$(ii)$,and $(iii)$,we get:$\frac{b}{c} + \frac{c}{a} + \frac{a}{b} = [\cos(\beta - \gamma) + \cos(\gamma - \alpha) + \cos(\alpha - \beta)] + i[\sin(\beta - \gamma) + \sin(\gamma - \alpha) + \sin(\alpha - \beta)] = 1$.Since $1 = 1 + 0i$,by equating the real parts,we get:$\cos(\beta - \gamma) + \cos(\gamma - \alpha) + \cos(\alpha - \beta) = 1$.

If $a = \cos \alpha + i\sin \alpha$,$b = \cos \beta + i\sin \beta$,$c = \cos \gamma + i\sin \gamma$ and $\frac{b}{c} + \frac{c}{a} + \frac{a}{b} = 1$,then $\cos (\beta - \gamma ) + \cos (\gamma - \alpha ) + \cos (\alpha - \beta )$ is equal to

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