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(C) The slopes of the lines are $m_1 = 	an(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$ and $m_2 = 	an(\frac{5\pi}{6}) = -\frac{1}{\sqrt{3}}$.Since the line

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$x^2-3 y^2=0$

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(C) The slopes of the lines are $m_1 = 	an(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$ and $m_2 = 	an(\frac{5\pi}{6}) = -\frac{1}{\sqrt{3}}$.Since the lines pass through the origin,their equations are $y = m_1 x$ and $y = m_2 x$.Substituting the slopes,we get $y = \frac{1}{\sqrt{3}}x$ and $y = -\frac{1}{\sqrt{3}}x$.This simplifies to $x - \sqrt{3}y = 0$ and $x + \sqrt{3}y = 0$.The combined equation is $(x - \sqrt{3}y)(x + \sqrt{3}y) = 0$.Using the identity $(a-b)(a+b) = a^2 - b^2$,we get $x^2 - (\sqrt{3}y)^2 = 0$.Therefore,the combined equation is $x^2 - 3y^2 = 0$.

The combined equation of the lines whose inclinations are $\frac{\pi}{6}$ and $\frac{5 \pi}{6}$,and passing through the origin,is

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