(D) Let $x = \sec \theta$. Since $x > 1$,we have $0 < \theta < \frac{\pi}{2}$.Then,$\sqrt{x^2-1} = \sqrt{\sec^2 \theta - 1} = \sqrt{\tan^2 \theta} = \

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(D) Let $x = \sec \theta$. Since $x > 1$,we have $0 Then,$\sqrt{x^2-1} = \sqrt{\sec^2 \theta - 1} = \sqrt{\tan^2 \theta} = \tan \theta$.Substituting this into the expression,we get $\cot ^{-1}\left(\frac{1}{\tan \theta}\right) = \cot ^{-1}(\cot \theta)$.Since $0 Substituting back $\theta = \sec ^{-1} x$,we get the simplest form as $\sec ^{-1} x$.