Question

Let $-\frac{\pi }{6}<\theta <-\frac{\pi }{12}$. Suppose ${\alpha }_1$ and ${\beta }_1$ 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:

• A. $2(\sec \theta -\tan \theta )$
• B. $2\sec \theta$
• C. $-2\tan \theta$
• D. $0$

Answer 1

The correct option is C $-2\tan \theta$

For $x^2-2x\sec \theta +1=0$
$x=\frac{2\sec \theta \pm \sqrt{4{\sec }^2\theta -4}}{2}=\sec \theta \pm \tan \theta$

For $x^2-2x\tan \theta -1=0$
$x=\frac{-2\tan \theta \pm \sqrt{4{\tan }^2\theta +4}}{2}=-\tan \theta \pm \sec \theta$

$\because -\frac{\pi }{6}<\theta <\frac{\pi }{12}$ and ${\alpha }_1>{\beta }_2$

$\Rightarrow {\alpha }_1=\sec \theta -\tan \theta$ and
${\beta }_2=-\tan \theta -\sec \theta$

${\alpha }_1+{\beta }_2=-2\tan \theta$