Mathematics - Quadratic Equation Question with Solution | TestHub

As ${\alpha }_1>{\beta }_1$ so $\text{'}+\text{'}$ sign will be used for ${\alpha }_1$, ${\beta }_2<{\alpha }_2$ so $\text{'}-\text{'}$ sign for ${\beta }_2$.
${\alpha }_1=\frac{2\sec \theta \pm \sqrt{4{\sec }^2 \theta -4}}{2},$ ${\beta }_2=\frac{-2\tan \mathrm{\theta }- \sqrt{4{\tan }^2 \theta +4}}{2}$
${\alpha }_1=\sec \theta +\left|\tan \theta \right|$  
${\beta }_2= -\tan \theta -\sec \theta$ 
${\alpha }_1=\sec \theta -\tan \theta$ $\left\{\because \theta \in \left(-\frac{\pi }{6}, -\frac{\pi }{12}\right)\right\}$  

So, $\mathit{tan}\mathit{ }\theta$ is $-ve$ $\&$ $\mathit{sec}\mathit{ }\theta$ is$+ve$
${\beta }_2=-tan \theta -sec \theta$
${\alpha }_1+{\beta }_2= -2\tan \theta$