If m_{1} and m_{2} are slopes of the lines re | vedclass

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(B) The given equation is of the form $Ax^{2} + 2Hxy + By^{2} = 0$,where $A = \sec^{2} 	heta - \sin^{2} 	heta$,$2H = -2 	an 	heta$,and $B = \sin^{

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$2$

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(B) The given equation is of the form $Ax^{2} + 2Hxy + By^{2} = 0$,where $A = \sec^{2} 	heta - \sin^{2} 	heta$,$2H = -2 	an 	heta$,and $B = \sin^{2} 	heta$. For a pair of lines $Ax^{2} + 2Hxy + By^{2} = 0$,the sum of slopes is $m_{1} + m_{2} = -\frac{2H}{B}$ and the product of slopes is $m_{1}m_{2} = \frac{A}{B}$. Here,$m_{1} + m_{2} = \frac{2 	an 	heta}{\sin^{2} 	heta}$ and $m_{1}m_{2} = \frac{\sec^{2} 	heta - \sin^{2} 	heta}{\sin^{2} 	heta}$. We know that $|m_{1} - m_{2}| = \sqrt{(m_{1} + m_{2})^{2} - 4m_{1}m_{2}}$. Substituting the values: $|m_{1} - m_{2}| = \sqrt{\left(\frac{2 	an 	heta}{\sin^{2} 	heta}ight)^{2} - 4\left(\frac{\sec^{2} 	heta - \sin^{2} 	heta}{\sin^{2} 	heta}ight)}$ $|m_{1} - m_{2}| = \sqrt{\frac{4 	an^{2} 	heta - 4(\sec^{2} 	heta - \sin^{2} 	heta) \sin^{2} 	heta}{\sin^{4} 	heta}}$ Using $	an^{2} 	heta = \frac{\sin^{2} 	heta}{\cos^{2} 	heta}$ and $\sec^{2} 	heta = \frac{1}{\cos^{2} 	heta}$: $|m_{1} - m_{2}| = \sqrt{\frac{4 \frac{\sin^{2} 	heta}{\cos^{2} 	heta} - 4(\frac{1}{\cos^{2} 	heta} - \sin^{2} 	heta) \sin^{2} 	heta}{\sin^{4} 	heta}} = \sqrt{\frac{4 \frac{\sin^{2} 	heta}{\cos^{2} 	heta} - 4 \frac{\sin^{2} 	heta}{\cos^{2} 	heta} + 4 \sin^{4} 	heta}{\sin^{4} 	heta}} = \sqrt{\frac{4 \sin^{4} 	heta}{\sin^{4} 	heta}} = \sqrt{4} = 2$.

If $m_{1}$ and $m_{2}$ are slopes of the lines represented by $(\sec^{2} \theta - \sin^{2} \theta) x^{2} - 2 \tan \theta xy + \sin^{2} \theta y^{2} = 0$,then $|m_{1} - m_{2}| = $

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