Show that the two straight lines $x^2(\tan^2 (\theta)+\cos^2 (\theta))-2xy\tan (\theta)+y^2.\sin^2 (\theta)=0$ make with x axis such that the difference of their tangents is $2$.
My Attempt: $$x^2(\tan^2 (\theta) +\cos^2 (\theta))-2xy\tan (\theta) + y^2 \sin^2 (\theta)=0$$
Let $y-m_1x=0$ and $y-m_2x=0$ be the two lines represented by the above equation. Their combined equation is: $$(y-m_1x)(y-m_2x)=0$$ $$y^2-(m_1+m_2)xy+(m_1m_2)x^2=0$$
How do I proceed further?
We have $$\left(\dfrac yx\right)^2\sin^2\theta-\dfrac yx(2\tan\theta)+\tan^2\theta+\cos^2\theta=0$$
If $m_1,m_2$ are the two roots
$m_1m_2=\dfrac{\tan^2\theta+\cos^2\theta}{\sin^2\theta}=\dfrac{s^2+c^4}{c^2s^2}$
$m_1+m_2=\dfrac{2s}{cs^2}=\dfrac2{cs}$
$$(m_1-m_2)^2=(m_1+m_2)^2-4m_1m_2=4\cdot\dfrac{1-(s^2+c^4)}{c^2s^2}=\dfrac{4c^2(1-c^2)}{c^2s^2}=4$$