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

Required value $=\tan {\theta }_1-\tan {\theta }_2$
$=m_1-m_2=\frac{2\sqrt{h^2-ab}}{\left|b\right|}$ 
Where $h=-\tan \alpha$

$\begin{array}{l}a={\tan }^2\alpha +{\cos }^2\alpha \\ b={\sin }^2\alpha \end{array}$

$=\frac{2\sqrt{{\tan }^2\alpha -\left({\tan }^2\alpha +{\cos }^2\alpha \right){\sin }^2\alpha }}{\left|{\sin }^2\alpha \right|}$

$=\frac{2.{\sin }^2\alpha }{{\sin }^2\alpha }=2$