Explanation

(a)
The given equation is
$(a - 2b + c)x^{2} + (b - 2c + a)x + (c - 2a + b) = 0$
$\because\sum(a - 2b + c) = 0$
$\therefore\$One root of this equation is 1
Now, $\sec{\theta + \tan{\theta = 1\ \ \ \ldots(i)}}$
We know that, $\sec^{2}{\theta - \tan^{2}{\theta = 1}}$
$\Rightarrow \sec{\theta - \tan{\theta = 1}}$ ….(ii)
On solving Eqs.(i) and (ii), we get
$\sec{\theta = 1}$
$\therefore$ One root of given equation is $\sec\theta$