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

If $a+b+c=0 ofa{x}^2+bx+c=0 then one of root is 1$

$\sec \theta +\tan \theta =1;\theta =0$ is the solution of the above equation.

$\therefore \sec 0°=1$ is a root of the

eq. $(a-2b+c)x^2+(b-2c+a)x+(c-2a+b)=0$

$\therefore \sec \theta$ in a root of the given equation.