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(D) Let $\alpha = \cos^{-1} \frac{1}{5\sqrt{2}}$. Then $\cos \alpha = \frac{1}{5\sqrt{2}}$.Since $	an^2 \alpha = \sec^2 \alpha - 1 = (5\sqrt{2})^2 -

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$3/29$

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(D) Let $\alpha = \cos^{-1} \frac{1}{5\sqrt{2}}$. Then $\cos \alpha = \frac{1}{5\sqrt{2}}$.Since $	an^2 \alpha = \sec^2 \alpha - 1 = (5\sqrt{2})^2 - 1 = 50 - 1 = 49$,we have $	an \alpha = 7$.Thus,$\alpha = 	an^{-1} 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 $\cos \beta = \frac{1}{\sqrt{17}}$.Then $	an \beta = \frac{\sin \beta}{\cos \beta} = \frac{4/\sqrt{17}}{1/\sqrt{17}} = 4$.Thus,$\beta = 	an^{-1} 4$.The expression becomes $	an(\alpha - \beta) = 	an(	an^{-1} 7 - 	an^{-1} 4)$.Using the formula $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$,we get:$	an(	an^{-1} 7 - 	an^{-1} 4) = \frac{7 - 4}{1 + (7)(4)} = \frac{3}{1 + 28} = \frac{3}{29}$.

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

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