Show that the two straight lines $\;$ $x^2 \left(\tan^2 \theta + \cos^2 \theta\right) - 2 xy \tan \theta + y^2 \sin^2 \theta = 0$ $\;$ make with the $X$ axis angles such that the difference of their tangents is $2$.
Equation of given pair of lines is:
$x^2 \left(\tan^2 \theta + \cos^2 \theta\right) - 2 xy \tan \theta + y^2 \sin^2 \theta = 0$ $\;\;\; \cdots \; (1a)$
Equation $(1a)$ can be written as
$x^2 \left(\dfrac{\tan^2 \theta + \cos^2 \theta}{\sin^2 \theta}\right) - 2 \dfrac{\tan \theta}{\sin^2 \theta} xy + y^2 = 0$ $\;\;\; \cdots \; (1b)$
Let the lines represented by equation $(1b)$ make angles $\phi_1$ and $\phi_2$ with the $X$ axis.
Then the combined equation of the lines given by equation $(1b)$ can be written as
$\left(y - x \tan \phi_1\right) \left(y - x \tan \phi_2\right) = 0$
i.e. $\;$ $y^2 - xy \tan \phi_2 - xy \tan \phi_1 + x^2 \tan \phi_1 \tan \phi_2 = 0$
i.e. $\;$ $y^2 - xy \left(\tan \phi_1 + \tan \phi_2\right) + x^2 \tan \phi_1 \tan \phi_2 = 0$ $\;\;\; \cdots \; (2)$
Since equations $(1b)$ and $(2)$ are identical equations, we have,
$\tan \phi_1 \tan \phi_2 = \dfrac{\tan^2 \theta + \cos^2 \theta}{\sin^2 \theta}$ $\;\;\; \cdots \; (3a)$ $\;$ and
$\tan \phi_1 + \tan \phi_2 = \dfrac{2 \tan \theta}{\sin^2 \theta}$ $\;\;\;\; \cdots \; (3b)$
Difference of tangents of the angles made by the lines given by equation $(1a)$ is $= \tan \phi_1 - \tan \phi_2$
Now,
$\begin{aligned}
\left(\tan \phi_1 - \tan \phi_2\right)^2 & = \left(\tan \phi_1 + \tan \phi_2\right)^2 - 4 \tan \phi_1 \tan \phi_2 \\\\
& = \left(\dfrac{2 \tan \theta}{\sin^2 \theta}\right)^2 - 4 \times \left(\dfrac{\tan^2 \theta + \cos^2 \theta}{\sin^2 \theta}\right) \\\\
& = \dfrac{4 \tan^2 \theta}{\sin^4 \theta} - 4 \times \left(\dfrac{\tan^2 \theta \sin^2 \theta + \sin^2 \theta \cos^2 \theta}{\sin^4 \theta}\right) \\\\
& = \dfrac{4 \tan^2 \theta \left(1 - \sin^2 \theta\right) - 4 \sin^2 \theta \cos^2 \theta}{\sin^4 \theta} \\\\
& = \dfrac{4 \tan^2 \theta \cos^2 \theta - 4 \sin^2 \theta \cos^2 \theta}{\sin^2 \theta} \\\\
& = \dfrac{4 \cos^2 \theta \left(\tan^2 \theta - \sin^2 \theta\right)}{\sin^4 \theta} \\\\
& = \dfrac{4 \cos^2 \theta \left(\dfrac{\sin^2 \theta - \sin^2 \theta \cos^2 \theta}{\cos^2 \theta}\right)}{\sin^4 \theta} \\\\
& = \dfrac{4 \sin^2 \theta \left(1 - \cos^2 \theta\right)}{\sin^4 \theta} \\\\
& = \dfrac{4 \times \sin^2 \theta \times \sin^2 \theta}{\sin^4 \theta}
\end{aligned}$
i.e. $\;$ $\left(\tan \phi_1 - \tan \phi_2\right)^2 = 4$
$\implies$ $\tan \phi_1 - \tan \phi_2 = 2$