1. Let $y = \cot^{-1}x$

   Then the inequality becomes $y^2 - 7y + 10 > 0$

2. Factorize the quadratic $y^2 - 7y + 10 = (y-5)(y-2)$

   So we need $(y-5)(y-2) > 0$

3. Solve the quadratic inequality in $y$

   Since the parabola opens upward, the product is positive outside the roots: $y<2 \quad \text{or} \quad y>5$

4. Use the principal value range of $\cot^{-1}x$

   For JEE, $\cot^{-1}x \in (0,\pi)$

   Hence $0<y<\pi$

   Now compare with the solution $y<2$ or $y>5$.

   Since $\pi \approx 3.14 < 5,$ the condition $y>5$ is impossible.

   Therefore only $0<y<2$ remains.

5. Convert back to $x$

   We have $\cot^{-1}x < 2$

   Since $\cot y$ is strictly decreasing on $(0,\pi)$, from $y<2$ we get $x = \cot y > \cot 2$

   Thus the solution set is $x \in (\cot 2,\infty)$

6. Match with options

   This is Option A.

Verification with stored answer

Stored correct answer: A

Our derived answer: A

So the answer agrees with the stored answer.