If ${\theta }_{1}and{\theta }_2$ are the angles made by the lines $\left(x^2+y^2\right)\left({\cos }^2\theta {\sin }^2\alpha +{\sin }^2\theta \right)={\left(x\tan \alpha -y\sin \theta \right)}^2$with positive axis of X in the anticlock wise direction, If $\theta =\frac{\pi }{6},\tan {\theta }_1+\tan {\theta }_{2}Equals$

Given equation$\begin{array}{l}\left(x^2+y^2\right)\left({\cos }^2\theta {\sin }^2\alpha +{\sin }^2\theta \right)={\left(x\tan \alpha -y\sin \theta \right)}^2\\ \Rightarrow \left({\cos }^2\theta {\sin }^2\alpha +{\sin }^2\theta -{\mathrm{Tan}}^2\alpha \right)x^2+2xy\mathrm{Tan}\alpha \sin \theta +\left({\cos }^2\theta {\sin }^2\alpha +{\sin }^2\theta -{\sin }^2\theta \right)y^2=0\\ \therefore \mathrm{Tan}{\theta }_1+\mathrm{Tan}{\theta }_2=m_1+m_2=\frac{-2\mathrm{Tan}\theta \sin \theta }{{\cos }^2\theta {\sin }^2\alpha +\cancel{{\sin }^2\theta }-\cancel{{\sin }^2\theta }}\\ \Rightarrow \frac{-\mathrm{Tan}\alpha }{\frac{3}{4}{\sin }^2\alpha }\end{array}$

$=-\frac{4}{3}\frac{\sin \alpha }{\cos \alpha }\times \frac{1}{{\sin }^2\alpha }\Rightarrow \frac{-8}{3\left(2\sin \alpha \cos \alpha \right)}=\frac{-8}{3\sin 2\alpha }=-\frac{8}{3}\cos ec2\alpha$