Let $\frac{-\mathrm{\pi }}{6}<\mathrm{\theta }<\frac{-\mathrm{\pi }}{12}$, suppose ${\mathrm{\alpha }}_1\&{\mathrm{\beta }}_1$ are roots of the equation ${\mathrm{x}}^2-2\mathrm{xsec\theta }+1=0$. ${\alpha }_{2 } and {\beta }_2$  are roots of the equation ${\mathrm{x}}^2+2\mathrm{xtan\theta }-1=0$.  If ${\mathrm{\alpha }}_1>{\mathrm{\beta }}_1$ and ${\mathrm{\alpha }}_2>{\mathrm{\beta }}_2$, then 

As ${\mathrm{\alpha }}_1>{\mathrm{\beta }}_1$,so ‘+’ sign will be used for ${\mathrm{\alpha }}_1$,
${\mathrm{\beta }}_2<{\mathrm{a}}_2$ , so ‘-‘ sign for ${\mathrm{\beta }}_2$
${\mathrm{\alpha }}_1=\frac{2\mathrm{sec\theta }\pm \sqrt{4{\sec }^2\mathrm{\theta }-4}}{2},{\mathrm{\beta }}_2=\frac{-2\mathrm{tan\theta }-\sqrt{4{\tan }^2\mathrm{\theta }+4}}{2}$
${\mathrm{\alpha }}_1=\mathrm{sec\theta }+\left|\mathrm{tan\theta }\right|, {\mathrm{\beta }}_2=-\mathrm{tan\theta }-\mathrm{sec\theta }$
${\mathrm{\alpha }}_1=\mathrm{sec\theta }-\mathrm{tan\theta }[\therefore \mathrm{\theta }\in (\frac{-\mathrm{\pi }}{6},\frac{-\mathrm{\pi }}{12})]$
So, sec$\mathrm{\theta }$ is +ve & tan$\mathrm{\theta }$ is –ve
${\mathrm{\alpha }}_1+{\mathrm{\beta }}_2=-2\mathrm{tan\theta } {\mathrm{\alpha }}_1-{\mathrm{\beta }}_2=2\mathrm{sec\theta }$