Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

${\left({\cot }^{-1}x\right)}^2-7\left({\cot }^{-1}x\right)+10>0$ $\left({\cot }^{-1}x-5\right)\left({\cot }^{-1}-2\right)>0$ ${\cot }^{-1}x\in (-\infty ,2)\cup (5,\infty )$ But ${\cot }^{-1}x$ lies in $(0,\pi )$ Now, from equation (1) ${\cot }^{-1}x\in (0,2)$ Now, it is clear from the graph $x\in (\cot 2,\infty )$