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(D) The equation of the pair of lines is $x^2 - 2xy 	an 	heta - y^2 = 0$.Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we get $a = 1

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$	an^{-1}(-2)$

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(D) The equation of the pair of lines is $x^2 - 2xy 	an 	heta - y^2 = 0$.Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we get $a = 1$,$2h = -2 	an 	heta$,and $b = -1$.The sum of the slopes $m_1 + m_2$ is given by the formula $\frac{-2h}{b}$.Substituting the values,we have $m_1 + m_2 = \frac{-(-2 	an 	heta)}{-1} = -2 	an 	heta$.Given that the sum of the slopes is $4$,we set $-2 	an 	heta = 4$.This simplifies to $	an 	heta = -2$.Therefore,$	heta = 	an^{-1}(-2)$.

If the sum of the slopes of the lines represented by $x^2 - 2xy \tan \theta - y^2 = 0$ is $4$,then $\theta =$

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