The solution of the inequality (cot^{-1}x)^2 | vedclass

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(B) Let $y = \cot^{-1}x$. The inequality becomes $y^2 - 5y + 6 > 0$. Factoring the quadratic expression,we get $(y - 3)(y - 2) > 0$. This inequality h

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$(-\infty, \cot 3) \cup (\cot 2, \infty)$

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(B) Let $y = \cot^{-1}x$. The inequality becomes $y^2 - 5y + 6 > 0$. Factoring the quadratic expression,we get $(y - 3)(y - 2) > 0$. This inequality holds when $y  3$. Substituting $y = \cot^{-1}x$,we have $\cot^{-1}x  3$. Since the function $\cot^{-1}x$ is a strictly decreasing function,the inequality sign reverses when we apply the cotangent function to both sides. For $\cot^{-1}x  \cot 2$. For $\cot^{-1}x > 3$,we get $x Combining these,the solution set is $x \in (-\infty, \cot 3) \cup (\cot 2, \infty)$.

The solution of the inequality $(\cot^{-1}x)^2 - 5\cot^{-1}x + 6 > 0$ is:

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