Animated Solution for Mathematics - Inverse Trigonometric Functions: All $x$ satisfying the inequality $({\cot }^{-1}x{)}^2-7({\cot }^{-1}x)+10>0$, lie in the interval:

When you first look at the inequality $({\cot }^{-1}x{)}^2-7({\cot }^{-1}x)+10>0$, it is natural to feel a bit overwhelmed. You see inverse trigonometric functions, squares, and a quadratic structure.

However, most complex-looking problems are just simple concepts wearing a mask. Let us perform a substitution by letting $t={\cot }^{-1}x$.

The problem transforms into a standard quadratic inequality:

We factorize this expression as:

Using the wavy curve method, we immediately see that this inequality holds true when $t<2$ or $t>5$.

Where Most Students Stumble

We are not solving for $t$; we are solving for $x$. Substituting back, we obtain:

This is where the trap lies. Many students rush to solve for $x$ without considering the domain and range of the inverse trigonometric functions.

Recall that the principal range of ${\cot }^{-1}x$ is $(0,\pi )$. Since $\pi \approx 3.14$, the maximum value ${\cot }^{-1}x$ can ever achieve is just under $3.14$.

Therefore, the condition ${\cot }^{-1}x>5$ is physically impossible. It is a mathematical ghost, and we discard it entirely. We are left with only one valid condition:

We are almost at the finish line. To isolate $x$, we must apply the cotangent function to all parts of the inequality $0<{\cot }^{-1}x<2$.

Stop and think: is the cotangent function increasing or decreasing? If you visualize the graph of $y=\cot \theta$ in the interval $(0,\pi )$, you will see it is strictly decreasing.

This is the crucial moment. When you apply a strictly decreasing function to an inequality, the inequality signs must flip:

Since $\cot (0)$ approaches $\infty$, we obtain:

Reading this from right to left, we find $x>\cot (2)$. In interval notation, the final solution is: $(\cot 2,\infty )$