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Q)

If $x=y\cos\large\frac{2\pi}{3}$$=z\cos\large\frac{4\pi}{3}$ then $xy+yz+zx$ is equal to

$(a)\;-1\qquad(b)\;0\qquad(c)\;1\qquad(d)\;2$

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A)
$x=y\cos\large\frac{2\pi}{3}$$=z\cos\large\frac{4\pi}{3}$
$\Rightarrow x=\large\frac{-y}{2}=\frac{-z}{2}$
$\Rightarrow y=z=-2x$
$\therefore xy+yz+zx=x(-2x)+(-2x)(-2x)+(-2x)(x)$
$\qquad\qquad\qquad=-2x^2+4x^2-2x^2$
$\qquad\qquad\qquad=-4x^2+4x^2$
$\qquad\qquad\qquad=0$
Hence (b) is the correct option.
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