Step 1: Understanding the Concept:
The problem asks for the tangent of the difference of two inverse trigonometric angles.
We assign variables to the inverse functions to convert them into standard trigonometric ratios.
Step 2: Key Formula or Approach:
We use the substitution $\alpha = \cos^{-1} \frac{1}{5\sqrt{2}}$ and $\beta = \sin^{-1} \frac{4}{\sqrt{17}}$.
The target expression becomes $\tan(\alpha - \beta)$.
The key formula is the tangent difference identity:
\[ \tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} \]
Step 3: Detailed Explanation:
From $\alpha = \cos^{-1} \frac{1}{5\sqrt{2}}$, we have $\cos \alpha = \frac{1}{5\sqrt{2}}$.
Using the Pythagorean identity $\sin^2 \alpha + \cos^2 \alpha = 1$:
\[ \sin \alpha = \sqrt{1 - \left(\frac{1}{5\sqrt{2}}\right)^2} = \sqrt{1 - \frac{1}{50}} = \sqrt{\frac{49}{50}} = \frac{7}{5\sqrt{2}} \]
Then, $\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{7/5\sqrt{2}}{1/5\sqrt{2}} = 7$.
From $\beta = \sin^{-1} \frac{4}{\sqrt{17}}$, we have $\sin \beta = \frac{4}{\sqrt{17}}$.
Using the identity $\cos^2 \beta + \sin^2 \beta = 1$:
\[ \cos \beta = \sqrt{1 - \left(\frac{4}{\sqrt{17}}\right)^2} = \sqrt{1 - \frac{16}{17}} = \sqrt{\frac{1}{17}} = \frac{1}{\sqrt{17}} \]
Then, $\tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{4/\sqrt{17}}{1/\sqrt{17}} = 4$.
Substitute $\tan \alpha$ and $\tan \beta$ into the difference formula:
\[ \tan(\alpha - \beta) = \frac{7 - 4}{1 + (7)(4)} = \frac{3}{1 + 28} = \frac{3}{29} \]
Step 4: Final Answer:
The calculated value of the expression is $\frac{3}{29}$.