$$\text{If } \alpha > \beta > \gamma > 0,\text{ then the expression}\cot^{-1}\!\left\{\beta+\frac{(1+\beta^2)}{(\alpha-\beta)}\right\} + \cot^{-1}\!\left\{\gamma+\frac{(1+\gamma^2)}{(\beta-\gamma)}\right\} + \cot^{-1}\!\left\{\alpha+\frac{(1+\alpha^2)}{(\gamma-\alpha)}\right\}\text{ is equal to:}$$

Using the identity $$\cot^{-1}(x) = \tan^{-1}(1/x)$$, the first term is:

$$\cot^{-1} \left( \frac{\beta(\alpha-\beta) + 1 + \beta^2}{\alpha-\beta} \right) = \cot^{-1} \left( \frac{\alpha\beta - \beta^2 + 1 + \beta^2}{\alpha-\beta} \right) = \tan^{-1} \left( \frac{\alpha-\beta}{1+\alpha\beta} \right)$$

This simplifies to: $$\tan^{-1}\alpha - \tan^{-1}\beta$$.

Similarly:

Term 2: $$\tan^{-1}\beta - \tan^{-1}\gamma$$

Term 3: $$\cot^{-1} \left( \frac{\alpha\gamma - \alpha^2 + 1 + \alpha^2}{\gamma - \alpha} \right) = \cot^{-1} \left( \frac{1+\alpha\gamma}{\gamma-\alpha} \right)$$

Note: Since $$\gamma - \alpha < 0$$, $$\cot^{-1}(x)$$ where $$x$$ is negative is $$\pi + \tan^{-1}(1/x)$$.

Term 3 $$= \pi + \tan^{-1} \left( \frac{\gamma-\alpha}{1+\gamma\alpha} \right) = \pi + \tan^{-1}\gamma - \tan^{-1}\alpha$$.

Sum: $$(\tan^{-1}\alpha - \tan^{-1}\beta) + (\tan^{-1}\beta - \tan^{-1}\gamma) + (\pi + \tan^{-1}\gamma - \tan^{-1}\alpha) = \mathbf{\pi}$$ (Option A).