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 ${\alpha }_1>{\beta }_2$ , 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 $-2\tan \theta$

$-\frac{\pi }{6}<\theta <-\frac{\pi }{12}$

$\Rightarrow x^2-2x\sec \theta +1=0$

$x=\frac{2\sec \theta \pm \sqrt{4{\sec }^2\theta -4}}{2}$

$x=\frac{\sec \theta \pm \tan \theta }{1}$

$\Rightarrow x^2+2x\tan \theta -1=0$

$x=\frac{-2\tan \theta \pm \sqrt{4\tan \theta +4}}{2}$

$=\theta -\tan \theta \pm \sec \theta$

Let ${\alpha }_1=\sec \theta -\tan \theta$

${\beta }_1=\sec \theta +\tan \theta$

${\alpha }_2=\sec \theta -\tan \theta$

${\beta }_2=-\sec \theta -\tan \theta$

then, ${\alpha }_1=\underset{(+ve)}{\underset{}{\underset{(+ve)}{\sec \theta }-\underset{(-ve)}{\tan \theta }}}>{\beta }_2=\underset{less(+ve)}{\underset{}{\underset{(-ve)}{-\sec \theta }+\underset{(-ve)}{\tan \theta }}}$

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