Let - (π)/(6) β _ 1 and α _ 2 > β _ 2 , then α _ 1 + β _ 2 equals

As ${\alpha}_{1} > {\beta}_{1}$ so $' +'$ sign will be used for ${\alpha}_{1}$, ${\beta}_{2} < {\alpha}_{2}$ so $' -'$ sign for ${\beta}_{2}$.
${\alpha}_{1} = \dfrac{2sec\theta \pm \sqrt{4\sec^{2}\theta - 4}}{2},{\beta}_{2} = \dfrac{- 2tan\theta - \sqrt{4\tan^{2}\theta + 4}}{2}$
${\alpha}_{1} = sec\theta + \left|\tan\theta\right|$
${\beta}_{2} = - tan\theta - sec\theta$
${\alpha}_{1} = sec\theta - tan\theta \left\{∵ \theta \in \left(- \dfrac{\pi}{6}, - \dfrac{\pi}{12}\right)\right\}$

So, $tan\theta$ is $- ve sec\theta$ is$+ ve$
${\beta}_{2} = - tan\theta - sec\theta$
${\alpha}_{1} + {\beta}_{2} = - 2tan\theta$&