Prove that $\;$ $\tan^{-1} \left(\dfrac{m}{n}\right) - \tan^{-1} \left(\dfrac{m - n}{m + n}\right) = \dfrac{\pi}{4}$
$\begin{aligned} \text{LHS } = \tan^{-1} \left(\dfrac{m}{n}\right) - \tan^{-1} \left(\dfrac{m - n}{m + n}\right) & = \tan^{-1} \left[\dfrac{\dfrac{m}{n} - \left(\dfrac{m - n}{m + n}\right)}{1 + \dfrac{m}{n} \times \left(\dfrac{m - n}{m + n}\right)}\right] \\\\ & = \tan^{-1} \left[\dfrac{\dfrac{m^2 + m \cdot n - m \cdot n + n^2}{n \left(m + n\right)}}{\dfrac{m \cdot n + n^2 + m^2 - m \cdot n}{n \left(m + n\right)}}\right] \\\\ & = \tan^{-1} \left[\dfrac{m^2 + n^2}{n^2 + m^2}\right] \\\\ & = \tan^{-1} \left(1\right) \\\\ & = \dfrac{\pi}{4} = \text{ RHS} \;\;\; \text{ Hence proved} \end{aligned}$