Question

If $\tan \theta$ and $\sec \theta$ are the roots of the equation $a{x}^2+bx+c=0$,then find the value of sec$\theta$-tan$\theta$

Answer 1

Since, $\sec \theta ,\tan \theta$ are the roots of the given equation,

$\sec \theta +\tan \theta =-\frac{b}{a}$ and

$\sec \theta \tan \theta =\frac{c}{a}$

$(\sec \theta -\tan \theta {)}^2=(\sec \theta +\tan \theta {)}^2-4\sec \theta \tan \theta$

$(\sec \theta -\tan \theta {)}^2={(-\frac{b}{a})}^2-4\frac{c}{a}$

$\therefore (\sec \theta -\tan \theta {)}^2=\frac{b^2-4ac}{a^2}$

$\Rightarrow \sec \theta -\tan \theta =\pm \frac{\sqrt{b^2-4ac}}{a}$

OR

w.k.t, ${\sec }^2\theta -{\tan }^2\theta =1$

$(\sec \theta -\tan \theta )(\sec \theta +\tan \theta )=1$

$\sec \theta -\tan \theta =\frac{1}{\sec \theta +\tan \theta }$

$\therefore \sec \theta -\tan \theta =-\frac{a}{b}$