Discussion Forum

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 roots of the equation $x^{2}+2x\tan\theta-1=0$. If  $\alpha_{1}>\beta_{1}$  and $\alpha_{2}>\beta_{2}$  , $\alpha_{1}+\beta_{2}$ equals

Answer:

Explanation:

Here,  $x^{2}-2x\sec\theta+1=0$ has roots  $\alpha_{1}$ and $\beta_{1}$.

 $\therefore$     $\alpha_{1},\beta_{1}=\frac{2\sec \theta\pm \sqrt{4\sec^{2}\theta-4}}{2\times 1}$

$=\frac{2\sec \theta\pm 2|\tan\theta|}{2}$

Since, $\theta \epsilon(-\frac{\pi}{6},-\frac{\pi}{12}),$

i.e,    Θ ε IV   quadrant = $\frac{2\sec \theta\pm2\tan \theta}{2}$

$\therefore$          $\alpha_{1}=\sec \theta-\tan\theta$      and    $\beta_{1}=\sec \theta+\tan\theta$

 (as $\alpha_{1}$ > $\beta_{1}$ )

 and     $x^{2}+2x\tan \theta-1=0$ has roots  $\alpha_{2}$  and $\beta_{2}$.

 i.e. $\alpha_{2} , \beta_{2}= \frac{-2\tan \theta\pm\sqrt{4\tan^{2}\theta+4}}{2}$

 $\therefore$      $\alpha_{2} =-\tan\theta+\sec \theta$

  and   $\beta_{2} =-\tan\theta-\sec \theta$  [as   $\alpha_{2}>\beta_{2}$ ]

 Thus,  $\alpha_{1}+\beta_{2}=-2\tan\theta$