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(A) Let $\alpha = \cos^{-1} \frac{1}{5\sqrt{2}}$. Then $\cos \alpha = \frac{1}{5\sqrt{2}}$.Since $\sin^2 \alpha = 1 - \cos^2 \alpha = 1 - \frac{1}{50}

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$\frac{3}{29}$

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(A) Let $\alpha = \cos^{-1} \frac{1}{5\sqrt{2}}$. Then $\cos \alpha = \frac{1}{5\sqrt{2}}$.Since $\sin^2 \alpha = 1 - \cos^2 \alpha = 1 - \frac{1}{50} = \frac{49}{50}$,we have $\sin \alpha = \frac{7}{5\sqrt{2}}$.Thus,$	an \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{7/5\sqrt{2}}{1/5\sqrt{2}} = 7$.Let $\beta = \sin^{-1} \frac{4}{\sqrt{17}}$. Then $\sin \beta = \frac{4}{\sqrt{17}}$.Since $\cos^2 \beta = 1 - \sin^2 \beta = 1 - \frac{16}{17} = \frac{1}{17}$,we have $\cos \beta = \frac{1}{\sqrt{17}}$.Thus,$	an \beta = \frac{\sin \beta}{\cos \beta} = \frac{4/\sqrt{17}}{1/\sqrt{17}} = 4$.We need to find $	an(\alpha - \beta) = \frac{	an \alpha - 	an \beta}{1 + 	an \alpha 	an \beta}$.Substituting the values,$	an(\alpha - \beta) = \frac{7 - 4}{1 + (7)(4)} = \frac{3}{1 + 28} = \frac{3}{29}$.

Considering only the principal values of the inverse trigonometric function,the value of $\tan \left(\cos ^{-1} \frac{1}{5 \sqrt{2}}-\sin ^{-1} \frac{4}{\sqrt{17}}\right)$ is

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