(A) Let $\cos ^{ - 1}x = \theta$. Then $x = \cos \theta$.Since $\sin^2 \theta + \cos^2 \theta = 1$,we have $\sin \theta = \sqrt{1 - \cos^2 \theta} = \

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(A) Let $\cos ^{ - 1}x = \theta$. Then $x = \cos \theta$.Since $\sin^2 \theta + \cos^2 \theta = 1$,we have $\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - x^2}$.Now,$\tan \theta = \frac{\sin \theta}{\cos \theta}$.Substituting the values,we get $\tan \theta = \frac{\sqrt{1 - x^2}}{x}$.Therefore,$\tan (\cos ^{ - 1}x) = \frac{\sqrt{1 - x^2}}{x}$.