If $\;$ $\tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \pi$, $\;$ then prove that $\;$ $x + y + z = x \cdot y \cdot z$
Let $\;$ $\tan^{-1} x = \alpha$, $\;$ $\tan^{-1} y = \beta$, $\;$ $\tan^{-1} z = \gamma$, $\;\;$ $\alpha, \; \beta, \; \gamma \in \left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$
Then, $\;$ $x = \tan \alpha$, $\;$ $y = \tan \beta$, $\;$ $z = \tan \gamma$
Given: $\;$ $\tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \pi$
i.e. $\;$ $\alpha + \beta + \gamma = \pi$
i.e. $\;$ $\alpha + \beta = \pi - \gamma$
i.e. $\;$ $\tan \left[\alpha + \beta\right] = \tan \left[\pi - \gamma\right]$
i.e. $\;$ $\dfrac{\tan \alpha + \tan \beta}{1 - \tan \alpha \times \tan \beta} = - \tan \gamma$
i.e. $\;$ $\dfrac{x + y}{1 - x \cdot y} = - z$
i.e. $\;$ $x + y = - z + x \cdot y \cdot z$
i.e. $\;$ $x + y + z = x \cdot y \cdot z$
Hence proved.