If $\sec \theta +\tan \theta =1$ , then root of the equation $\left(a-2b+c\right)x^2+\left(b-2c+a\right)x+\left(c-2a+b\right)=0$  is

$D=b^2-4ac$

${\left(b-2c+a\right)}^2-4\left(a-2b+c\right)\left(c-2a+b\right)$

$=b^2+4c^2+a^2-4bc-4ac+2ab-4$ $\left(ac-2a^2+ab-2bc+4ab-2b^2+c^2-2ac+bc\right)$

$=b^2+4c^2+a^2-4bc-4ac+4ab-4$ $\left(-ac-2a^2+5ab-bc-2b^2+c^2\right)$

$\Rightarrow b^2+4c^2+a^2-4bc-4ac+2ab+4ac+8a^2-20ab+4bc+8b^2-4c^2$

$\Rightarrow 9a^2+9b^2-18ab$

$\Rightarrow {\left(3a-3b\right)}^2$

$\therefore x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$=\frac{-\left(b-2c+a\right)\pm \sqrt{{\left(3a-3b\right)}^2}}{2\left(a-2b+c\right)}$

$=\frac{-\left(b-2c+a\right)\pm \left(3a-3b\right)}{2\left(a-2b+c\right)}$

$=\frac{-b+2c-a+3a-3b}{2\left(a-2b+c\right)}$

Or

$\frac{-b+2c-a-3a+3b}{2\left(a-2b+c\right)}$

$=\frac{2a-4b+2c}{2\left(a-2b+c\right)}$

Or

$\frac{2b-4a+2c}{2\left(a-2b+c\right)}$

$=\frac{2\left(a-2b+c\right)}{2\left(a-2b+c\right)}$

Or

$\frac{2\left(b-2a+c\right)}{2\left(a-2b+c\right)}$

$\sec \theta +\tan \theta =1$

${\sec }^2\theta ={\left(1-\tan \theta \right)}^2$

$1+{\tan }^2\theta =1-2\tan \theta +{\tan }^2\theta$

$\tan \theta =0$

$\therefore \sec \theta +\tan \theta =1$

$\therefore \sec \theta =1$