Let −$\frac{π}{6}$ < θ < − $\frac{π}{12}$. Suppose α₁ and β₁ are the roots of the equation x² − 2x sec θ + 1 = 0 and α₂ and β₂ are the roots of the equation x² + 2x tan θ − 1 = 0. If α₁ > β₁ and α₂ > β₂, then α₁ + β₂ equals:
−2 tan θ
For \(x^2-2x\sec\theta+1=0\), the roots are \(x = \sec\theta \pm \sqrt{\sec^2\theta-1} = \sec\theta \pm |\tan\theta|\).
Since \(-\pi/6<\theta<-\pi/12\), \(\theta\) lies in the fourth quadrant where \(\cos\theta>0\) (so \(\sec\theta>0\)) and \(\tan\theta<0\), hence \(|\tan\theta|=-\tan\theta\).
So the roots are \(\sec\theta-\tan\theta\) and \(\sec\theta+\tan\theta\); since \(\tan\theta<0\), the larger root is \(\alpha_1=\sec\theta-\tan\theta\) and \(\beta_1=\sec\theta+\tan\theta\).
For \(x^2+2x\tan\theta-1=0\), the roots are \(x=-\tan\theta\pm\sqrt{\tan^2\theta+1}=-\tan\theta\pm\sec\theta\).
Since \(\sec\theta>0\), the larger root is \(\alpha_2=\sec\theta-\tan\theta\) and the smaller root is \(\beta_2=-\tan\theta-\sec\theta\).
So \(\alpha_1+\beta_2 = (\sec\theta-\tan\theta)+(-\tan\theta-\sec\theta) = -2\tan\theta\).
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