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{1}{x} = \frac

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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, $\frac{1}{x} = \frac{\cos 	heta}{k}$, $\frac{1}{y} = \frac{\cos \left( 	heta + \frac{2\pi}{3} ight)}{k}$, and $\frac{1}{z} = \frac{\cos \left( 	heta + \frac{4\pi}{3} ight)}{k}$.Summing these, $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{k} \left[ \cos 	heta + \cos \left( 	heta + \frac{2\pi}{3} ight) + \cos \left( 	heta + \frac{4\pi}{3} ight) ight]$.Using the sum-to-product formula $\cos A + \cos B = 2 \cos \left( \frac{A+B}{2} ight) \cos \left( \frac{A-B}{2} ight)$ for the last two terms:$\cos \left( 	heta + \frac{2\pi}{3} ight) + \cos \left( 	heta + \frac{4\pi}{3} ight) = 2 \cos \left( \frac{2	heta + 2\pi}{2} ight) \cos \left( \frac{-2\pi/3}{2} ight) = 2 \cos (	heta + \pi) \cos \left( -\frac{\pi}{3} ight)$.Since $\cos (	heta + \pi) = -\cos 	heta$ and $\cos \left( -\frac{\pi}{3} ight) = \frac{1}{2}$, this becomes $2(-\cos 	heta) \left( \frac{1}{2} ight) = -\cos 	heta$.Substituting back: $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{k} [\cos 	heta - \cos 	heta] = 0$.

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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