If \[ f(x)= \begin{cases} \dfrac{x-2}{|x-2|}+a, & x<2, \\[6pt] a+b, & x=2, \\[6pt] \dfrac{x-2}{|x-2|}+b, & x>2, \end{cases} \] is continuous at \(x=2\), then
Simplify the inverse trigonometric expression to find its principal value: \[ \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{3}\right) \]