Solve: $\;$ $\tan \left\{2 \tan^{-1} \left(\dfrac{1}{5}\right) - \dfrac{\pi}{4}\right\}$
$\left[\text{Note: } 2 \tan^{-1} x = \tan^{-1} \left(\dfrac{2 x}{1 - x^2}\right), \;\;\; x^2 < 1 \right.$
$\left. \tan^{-1} x - \tan^{-1} y = \tan^{-1} \left(\dfrac{x - y}{1 + x \cdot y}\right) \right]$
$\tan \left\{2 \tan^{-1} \left(\dfrac{1}{5}\right) - \dfrac{\pi}{4}\right\}$
$= \tan \left\{\tan^{-1} \left(\dfrac{\dfrac{2}{5}}{1 - \dfrac{1}{25}}\right) - \tan^{-1} \left(1\right) \right\}$
$= \tan \left\{\tan^{-1} \left(\dfrac{5}{12}\right) - \tan^{-1} \left(1\right) \right\}$
$= \tan \left\{\tan^{-1} \left(\dfrac{\dfrac{5}{12} - 1}{1 + \dfrac{5}{12} \times 1}\right) \right\}$
$= \tan \left\{\tan^{-1} \left(- \dfrac{7}{17}\right) \right\}$
$= \tan \left\{- \tan^{-1} \left(\dfrac{7}{17}\right) \right\}$
$= - \tan \left\{\tan^{-1} \left(\dfrac{7}{17}\right)\right\}$
$= - \dfrac{7}{17}$