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

Given: Evaluate $E = \tan\left(\cos^{-1}\frac{1}{5\sqrt{2}} - \sin^{-1}\frac{4}{\sqrt{17}}\right)$.

Let $A = \cos^{-1}\frac{1}{5\sqrt{2}}$ and $B = \sin^{-1}\frac{4}{\sqrt{17}}$, so $E = \tan(A - B)$.

Since $\cos A = \frac{1}{5\sqrt{2}} > 0$ and $A$ lies in $[0, \pi]$, we have $\sin A > 0$ and can determine the opposite side: $\sqrt{(5\sqrt{2})^2 - 1^2} = 7$. Thus, $\tan A = 7$.

Similarly, $\sin B = \frac{4}{\sqrt{17}} > 0$ and $B$ lies in $[-\frac{\pi}{2}, \frac{\pi}{2}]$, so $\cos B > 0$ and the adjacent side is $\sqrt{(\sqrt{17})^2 - 4^2} = 1$. Therefore, $\tan B = 4$.

Apply the identity $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$.

$E = \frac{7 - 4}{1 + 7 \cdot 4} = \frac{3}{29}$

The value is $\frac{3}{29}$, so the answer is $\boxed{\text{C}}$.