Compute
\[\tan \left( 2 \arctan \frac{1}{5} - \frac{\pi}{4} \right).\]
To solve this problem, I need to use some trigonometric identities and inverse functions. First, I recall that the tangent of a difference of angles is given by \[\tan (A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}.\] I can use this formula to rewrite the given expression as \[\frac{\tan \left( 2 \arctan \frac{1}{5} \right) - \tan \left( \frac{\pi}{4} \right)}{1 + \tan \left( 2 \arctan \frac{1}{5} \right) \tan \left( \frac{\pi}{4} \right)}.\] Next, I need to simplify the term $\tan \left( 2 \arctan \frac{1}{5} \right)$. I recall another identity that says that \[\tan (2A) = \frac{2 \tan A}{1 - \tan^2 A}.\] I can apply this formula to the term $\tan \left( 2 \arctan \frac{1}{5} \right)$ by letting $A = \arctan \frac{1}{5}$. This gives me \[\tan \left( 2 \arctan \frac{1}{5} \right) = \frac{2 \tan \left( \arctan \frac{1}{5} \right)}{1 - \tan^2 \left( \arctan \frac{1}{5} \right)}.\] Now I need to simplify the term $\tan \left( \arctan \frac{1}{5} \right)$. I know that the inverse tangent function $\arctan$ is the angle whose tangent is the given value. That is, if $\arctan x = \theta$, then $\tan \theta = x$. So, $\tan \left( \arctan \frac{1}{5} \right)$ is just $\frac{1}{5}$. Simplifying this fraction, I get \[\tan \left( 2 \arctan \frac{1}{5} \right) = \frac{2}{5} \cdot \frac{25}{24} = \frac{5}{12}.\] Finally, I need to simplify the term $\tan \left( \frac{\pi}{4} \right)$. Substituting this into the expression, I get \[\frac{\tan \left( 2 \arctan \frac{1}{5} \right) - \tan \left( \frac{\pi}{4} \right)}{1 + \tan \left( 2 \arctan \frac{1}{5} \right) \tan \left( \frac{\pi}{4} \right)} = \frac{\frac{5}{12} - 1}{1 + \frac{5}{12} \cdot 1}.\] Simplifying this fraction, I get \[\frac{\frac{5}{12} - 1}{1 + \frac{5}{12} \cdot 1} = \frac{-\frac{7}{12}}{\frac{17}{12}} = -\frac{7}{17}.\]

