If x₁, x₂, x₃,…., x_n  are positive numbers in decreasing order then $\cot^{-1}(\frac{1+x_1x_2}{x_1-x_2})+cot^{-1}(\frac{1+x_2x_3}{x_2-x_3})+.....+cot^{-1}(\frac{1+x_{n-1}-x_n}{x_{n-1}-x_n})+cot^{-1}(\frac{1+x_nx_1}{x_n-x_1})$ is equal to

π

The given expression is equal to

$\tan^{-1}(\frac{x_1-x_2}{1+x_1x_2})+tan^{-1}(\frac{x_2-x_3}{1+x_2x_3})+.....+tan^{-1}(\frac{x_{n-1}-x_{n-2}}{1+x_nx_{n-1}})+π-cot^{-1}(-\frac{1+x_nx_1}{x_1-x_n})$

= tan^-1x₁ - tan^-1x₂ + tan^-1x₂ - tan^-1x₁ + … + tan^-1x_n-1- tan^-1x_n + π - tan^–1$(\frac{x_1-x_n}{1+x_nx_n})$.

= tan^-1x₁ - tan^-1x_n + π- (tan^-1x₁ - tan^-1x_n) = π.

Hence (B) is the correct answer.