If frac{x}{cos theta} = frac{y}{cos left( the | vedclass

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(B) Let $\frac{x}{\cos 	heta} = \frac{y}{\cos \left( 	heta - \frac{2\pi}{3} ight)} = \frac{z}{\cos \left( 	heta + \frac{2\pi}{3} ight)} = k.$Th

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(B) Let $\frac{x}{\cos 	heta} = \frac{y}{\cos \left( 	heta - \frac{2\pi}{3} ight)} = \frac{z}{\cos \left( 	heta + \frac{2\pi}{3} ight)} = k.$Then,$x = k \cos 	heta, y = k \cos \left( 	heta - \frac{2\pi}{3} ight),$ and $z = k \cos \left( 	heta + \frac{2\pi}{3} ight).$Now,$x + y + z = k \left[ \cos 	heta + \cos \left( 	heta - \frac{2\pi}{3} ight) + \cos \left( 	heta + \frac{2\pi}{3} ight) ight].$Using the formula $\cos(A - B) + \cos(A + B) = 2 \cos A \cos B,$$x + y + z = k \left[ \cos 	heta + 2 \cos 	heta \cos \left( \frac{2\pi}{3} ight) ight].$Since $\cos \left( \frac{2\pi}{3} ight) = -\frac{1}{2},$$x + y + z = k \left[ \cos 	heta + 2 \cos 	heta \left( -\frac{1}{2} ight) ight] = k [\cos 	heta - \cos 	heta] = k(0) = 0.$Therefore,$x + y + z = 0.$

If $\frac{x}{\cos \theta} = \frac{y}{\cos \left( \theta - \frac{2\pi}{3} \right)} = \frac{z}{\cos \left( \theta + \frac{2\pi}{3} \right)},$ then $x + y + z = $

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