Let $-\frac{\pi}{6}<\theta<-\frac{\pi}{12}$. Suppose $\theta_{1}$ and $B_{1}$ 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 $x^{2}+2x\tan\theta-1=0$. If $\alpha_{1}>\beta_{1}$ and $\alpha_{2}>\beta_{2}$, then $\alpha_{1}+\beta_{2}$ equals

Let -π/6<θ<-π/12. Suppose θ₁ and B₁ are the roots of the equation x²-2xsecθ+1=0 and α₂ and β₂ are the roots of the equation x²+2xtanθ-1=0. If α₁>β₁ and α₂>β₂, then α₁+β₂ equals $-\dfrac{\pi}{6} < \theta < -\dfrac{\pi}{12}$ $\text{i.e. } \theta \in (-30^\circ, -15^\circ)$ $x^{2}-2x\sec\theta+1=0 \to \alpha_{1},\ \beta_{1}$ $x^{2}+2x\tan\theta-1=0 \to \alpha_{2},\ \beta_{2}$ $\text{Find } \alpha_{1}+\beta_{2}$ $x = \dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ $\tan\theta < 0,\quad \sec\theta > 0 \text{ for this } \theta$ $x = \dfrac{2\sec\theta \pm \sqrt{4\sec^{2}\theta-4}}{2}$ $= \sec\theta \pm \sqrt{\sec^{2}\theta-1}$ $\sec^{2}\theta-1 = \tan^{2}\theta$ $x = \sec\theta \pm |\tan\theta|$ $|\tan\theta| = -\tan\theta$ $x = \sec\theta - \tan\theta \ \text{ or } \ \sec\theta + \tan\theta$ $\alpha_{1} = \sec\theta - \tan\theta,\quad \beta_{1} = \sec\theta + \tan\theta$ $\alpha_{1}+\beta_{1} = 2\sec\theta,\quad \alpha_{1}\beta_{1} = \sec^{2}\theta-\tan^{2}\theta = 1$ $x = \dfrac{-2\tan\theta \pm \sqrt{4\tan^{2}\theta+4}}{2}$ $= -\tan\theta \pm \sqrt{\tan^{2}\theta+1}$ $\tan^{2}\theta+1 = \sec^{2}\theta$ $x = -\tan\theta \pm |\sec\theta|$ $|\sec\theta| = \sec\theta$ $x = \sec\theta - \tan\theta \ \text{ or } \ -\tan\theta - \sec\theta$ $\alpha_{2} = \sec\theta - \tan\theta,\quad \beta_{2} = -\tan\theta - \sec\theta$ $\alpha_{2}+\beta_{2} = -2\tan\theta,\quad \alpha_{2}\beta_{2} = -1$ $\alpha_{1}+\beta_{2} = (\sec\theta-\tan\theta)+(-\tan\theta-\sec\theta)$ $= -2\tan\theta$ $\alpha_{1}+\beta_{2} = -2\tan\theta \ \Rightarrow\ \text{option (3)}$