If x cos theta = y cos left( theta + frac{2pi | vedclass

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(C) Let $x \cos 	heta = y \cos \left( 	heta + \frac{2\pi}{3} ight) = z \cos \left( 	heta + \frac{4\pi}{3} ight) = k$.Then,$\frac{k}{x} = \cos \

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(C) Let $x \cos 	heta = y \cos \left( 	heta + \frac{2\pi}{3} ight) = z \cos \left( 	heta + \frac{4\pi}{3} ight) = k$.Then,$\frac{k}{x} = \cos 	heta$,$\frac{k}{y} = \cos \left( 	heta + \frac{2\pi}{3} ight)$,and $\frac{k}{z} = \cos \left( 	heta + \frac{4\pi}{3} ight)$.Adding these three equations,we get:$\frac{k}{x} + \frac{k}{y} + \frac{k}{z} = \cos 	heta + \cos \left( 	heta + \frac{2\pi}{3} ight) + \cos \left( 	heta + \frac{4\pi}{3} ight)$.Using the sum-to-product formula $\cos A + \cos B = 2 \cos \left( \frac{A+B}{2} ight) \cos \left( \frac{A-B}{2} ight)$:$\cos \left( 	heta + \frac{2\pi}{3} ight) + \cos \left( 	heta + \frac{4\pi}{3} ight) = 2 \cos \left( \frac{2	heta + 2\pi}{2} ight) \cos \left( \frac{-2\pi/3}{2} ight) = 2 \cos (	heta + \pi) \cos \left( -\frac{\pi}{3} ight)$.Since $\cos (	heta + \pi) = -\cos 	heta$ and $\cos (-\pi/3) = 1/2$,the sum becomes:$2(-\cos 	heta)(1/2) = -\cos 	heta$.Thus,$\frac{k}{x} + \frac{k}{y} + \frac{k}{z} = \cos 	heta - \cos 	heta = 0$.Since $k 
eq 0$,we have $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$.

If $x \cos \theta = y \cos \left( \theta + \frac{2\pi}{3} \right) = z \cos \left( \theta + \frac{4\pi}{3} \right)$,then the value of $\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$ is equal to

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