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

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

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$xy + yz + zx = 0$

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(B) Let $x \sin 	heta = y \sin \left( 	heta + \frac{2\pi}{3} ight) = z \sin \left( 	heta + \frac{4\pi}{3} ight) = k$.Then $x = \frac{k}{\sin 	heta}$,$y = \frac{k}{\sin \left( 	heta + \frac{2\pi}{3} ight)}$,and $z = \frac{k}{\sin \left( 	heta + \frac{4\pi}{3} ight)}$.Consider the sum $xy + yz + zx = k^2 \left[ \frac{1}{\sin 	heta \sin \left( 	heta + \frac{2\pi}{3} ight)} + \frac{1}{\sin \left( 	heta + \frac{2\pi}{3} ight) \sin \left( 	heta + \frac{4\pi}{3} ight)} + \frac{1}{\sin \left( 	heta + \frac{4\pi}{3} ight) \sin 	heta} ight]$.Using the identity $\sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)]$,the denominator terms simplify.Alternatively,using the property that $\sin 	heta + \sin \left( 	heta + \frac{2\pi}{3} ight) + \sin \left( 	heta + \frac{4\pi}{3} ight) = 0$,we can show that $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$.Multiplying by $xyz$,we get $yz + zx + xy = 0$.

If $x \sin \theta = y \sin \left( \theta + \frac{2\pi}{3} \right) = z \sin \left( \theta + \frac{4\pi}{3} \right)$,then:

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