The difference of the tangents of the angles | vedclass

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(C) The given equation is $(	an ^2 \alpha+\cos ^2 \alpha) x^2 - (2 	an \alpha) xy + (\sin ^2 \alpha) y^2 = 0$. Dividing by $x^2$,we get $(\sin ^2 \a

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$2$

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(C) The given equation is $(	an ^2 \alpha+\cos ^2 \alpha) x^2 - (2 	an \alpha) xy + (\sin ^2 \alpha) y^2 = 0$. Dividing by $x^2$,we get $(\sin ^2 \alpha) m^2 - (2 	an \alpha) m + (	an ^2 \alpha + \cos ^2 \alpha) = 0$,where $m = y/x$. Let $m_1$ and $m_2$ be the roots of this quadratic equation. Then $m_1 + m_2 = \frac{2 	an \alpha}{\sin ^2 \alpha} = \frac{2}{\sin \alpha \cos \alpha}$ and $m_1 m_2 = \frac{	an ^2 \alpha + \cos ^2 \alpha}{\sin ^2 \alpha} = \sec^2 \alpha + \cot^2 \alpha$. The difference of the slopes is $|m_1 - m_2| = \sqrt{(m_1 + m_2)^2 - 4m_1 m_2}$. $|m_1 - m_2| = \sqrt{\frac{4}{\sin^2 \alpha \cos^2 \alpha} - 4 \left( \frac{	an^2 \alpha + \cos^2 \alpha}{\sin^2 \alpha} ight)}$. $|m_1 - m_2| = \sqrt{\frac{4 - 4 \cos^2 \alpha (	an^2 \alpha + \cos^2 \alpha)}{\sin^2 \alpha \cos^2 \alpha}} = \sqrt{\frac{4 - 4 \sin^2 \alpha - 4 \cos^4 \alpha}{\sin^2 \alpha \cos^2 \alpha}}$. Using $1 - \sin^2 \alpha = \cos^2 \alpha$,we get $\sqrt{\frac{4 \cos^2 \alpha - 4 \cos^4 \alpha}{\sin^2 \alpha \cos^2 \alpha}} = \sqrt{\frac{4 \cos^2 \alpha (1 - \cos^2 \alpha)}{\sin^2 \alpha \cos^2 \alpha}} = \sqrt{\frac{4 \cos^2 \alpha \sin^2 \alpha}{\sin^2 \alpha \cos^2 \alpha}} = \sqrt{4} = 2$.

The difference of the tangents of the angles which the lines $(\tan ^2 \alpha+\cos ^2 \alpha) x^2-2 x y \tan \alpha +(\sin ^2 \alpha) y^2=0$ make with the $X$-axis is

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