If $\;$ $a > b > c > 0$, $\;$ then prove that $\;$ $\cot^{-1} \left[\dfrac{a \cdot b + 1}{a - b}\right] + \cot^{-1} \left[\dfrac{b \cdot c + 1}{b - c}\right] + \cot^{-1} \left[\dfrac{c \cdot a + 1}{c - a}\right] = \pi$
Given: $\;$ $a > b \implies a - b > 0$; $\;$ $b > c \implies b - c > 0$; $\;$ $a > c \implies c - a < 0$
$\because$ $\;$ $a > 0, \; b > 0, \; c > 0$ $\;$ $\implies$ $a \cdot b > 0$, $\;$ $b \cdot c > 0$, $\;$ $c \cdot a > 0$
$\therefore \;$ $\cot^{-1} \left[\dfrac{a \cdot b + 1}{a - b}\right] + \cot^{-1} \left[\dfrac{b \cdot c + 1}{b - c}\right] + \cot^{-1} \left[\dfrac{c \cdot a + 1}{c - a}\right]$
$= \cot^{-1} \left[\dfrac{a \cdot b + 1}{a - b}\right] + \cot^{-1} \left[\dfrac{b \cdot c + 1}{b - c}\right] + \cot^{-1} \left[- \dfrac{c \cdot a + 1}{a - c}\right]$
$\left\{\text{Note: } \cot^{-1} \left(- x\right) = \pi - \cot^{-1} \left(x\right), \;\;\; x \in R \right\}$
$= \cot^{-1} \left[\dfrac{a \cdot b + 1}{a - b}\right] + \cot^{-1} \left[\dfrac{b \cdot c + 1}{b - c}\right] + \pi - \cot^{-1} \left[\dfrac{c \cdot a + 1}{a - c}\right]$
$\left\{\text{Note: } \cot^{-1} \left(x\right) - \cot^{-1} \left(y\right) = \cot^{-1} \left(\dfrac{x \cdot y + 1}{y - x}\right)\right\}$
$= \cot^{-1} \left(b\right) - \cot^{-1} \left(a\right) + \cot^{-1} \left(c\right) - \cot^{-1} \left(b\right) + \pi - \cot^{-1} \left(c\right) + \cot^{-1} \left(a\right)$
$= \pi$
Hence proved.