Step-by-step Solution:

\[ \begin{aligned} & \text{As } \cos \sec^{-1} \frac{5}{3} = \cot^{-1} \frac{4}{3}, \text{ the given sum is: } \\ & \cot \left[ \tan^{-1} \frac{3}{4} + \tan^{-1} \frac{2}{3} \right] \\ & = \cot \left[ \tan^{-1} \left[ \frac{\frac{3}{4} + \frac{2}{3}}{1 - \frac{3}{4} \cdot \frac{2}{3}} \right] \right] \\ & = \cot \left[ \cot^{-1} \left[ \frac{6}{17} \right] \right] = \frac{6}{17}. \end{aligned} \]