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

From the trigonometric equation sec theta + tan theta = 1, we use the standard identity (sec^2 theta - tan^2 theta) = 1, which factors as (sec theta - tan theta)(sec theta + tan theta) = 1. Substituting the given value yields sec theta - tan theta = 1, and solving the system of equations provides sec theta = 1 and tan theta = 0. Evaluating the coefficients of the quadratic equation gives (a - 2b + c) + (b - 2c + a) + (c - 2a + b) = 0, meaning the sum of the coefficients is zero. Therefore, x = 1 is a root of the quadratic equation, and since sec theta also equals 1, the root is sec theta.

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