$Let-\frac{\pi }{6}<\theta <-\frac{\pi }{12}.$ Suppose ${\alpha }_1$ and ${\beta }_1$ are the roots of the equation $x^2-2x\sec \theta +1=0,$ and ${\alpha }_2$ and ${\beta }_2$ are the roots of the equation $x^2+2x\tan \theta -1=0$, and $if{\alpha }_1>{\beta }_{1}and{\alpha }_2>{\beta }_2,then{\alpha }_1+{\beta }_2$ equals   
 

$Here,xH2-2x\sec \theta +1=0hasroots{\alpha }_{1}and{\beta }_1.$

$\therefore {\alpha }_1,{\beta }_1=\frac{2\sec \theta \pm \sqrt{4{\sec }^2\theta -4}}{2\times 1}=\frac{2\sec \theta \pm 2\left|\tan \theta \right|}{2}$

$Since,\theta \in (-\frac{\pi }{6},-\frac{\pi }{12}),i.e.\theta \in IVquadrant=\frac{2\sec \theta \mp 2ta\theta }{2}$

$\therefore {\alpha }_1=\sec \theta -\tan \theta and{\beta }_1=\sec \theta +\tan \theta [as{\alpha }_1>{\beta }_1]and{x}^2+2x\tan \theta -1=0hasroots{\alpha }_{2}and{\beta }_2.$

$i.e.{\alpha }_2,{\beta }_2=\frac{-2\tan \theta \pm \sqrt{4{\tan }^2\theta +4}}{2}, \therefore {\alpha }_2=-\tan \theta +\sec \theta$

$and{\beta }_2=-\tan \theta -\sec \theta [as{\alpha }_2>{\beta }_2]Thus,{\alpha }_1+{\beta }_2=-2\tan \theta$