(B) Given that $x = \sec \theta + \tan \theta$. We know that $\sec^2 \theta - \tan^2 \theta = 1$,which can be written as $(\sec \theta - \tan \theta)(

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(B) Given that $x = \sec \theta + \tan \theta$. We know that $\sec^2 \theta - \tan^2 \theta = 1$,which can be written as $(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1$. Therefore,$\frac{1}{x} = \frac{1}{\sec \theta + \tan \theta} = \sec \theta - \tan \theta$. Now,$x + \frac{1}{x} = (\sec \theta + \tan \theta) + (\sec \theta - \tan \theta)$. $x + \frac{1}{x} = 2 \sec \theta$.