(B) Let $x = \tan \theta$,where $\theta = \tan^{-1} x$.Then,$\sqrt{1 + x^2} = \sqrt{1 + \tan^2 \theta} = \sqrt{\sec^2 \theta} = \sec \theta$.Substitut

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(B) Let $x = \tan \theta$,where $\theta = \tan^{-1} x$.Then,$\sqrt{1 + x^2} = \sqrt{1 + \tan^2 \theta} = \sqrt{\sec^2 \theta} = \sec \theta$.Substituting this into the expression:$\tan \left[ \sec^{-1} \sqrt{1 + x^2} \right] = \tan \left[ \sec^{-1} (\sec \theta) \right]$.Since $\sec^{-1} (\sec \theta) = \theta$,we have:$\tan \theta = x$.Therefore,the correct option is $B$.