Question

The solution set of the inequality $\left({\cot }^{-1}x\right)\left({\tan }^{-1}x\right)+(2-\frac{\pi }{2}){\cot }^{-1}x-3{\tan }^{-1}x-3(2-\frac{\pi }{2})>0$ is

• A. $x\in (\tan 2,\tan 3)$
• B. $x\in (\cot 3,\cot 2)$
• C. $x\in (-\infty ,\tan 2)\cup (\tan 3,\infty )$
• D. $x\in (-\infty ,\cot 3)\cup (\cot 2,\infty )$

Answer 1

The correct option is B $x\in (\cot 3,\cot 2)$

$\left({\cot }^{-1}x\right)\left({\tan }^{-1}x\right)+(2-\frac{\pi }{2}){\cot }^{-1}x-3{\tan }^{-1}x-3(2-\frac{\pi }{2})>0$
$\Rightarrow {\cot }^{-1}x({\tan }^{-1}x+2-\frac{\pi }{2})-3({\tan }^{-1}x+2-\frac{\pi }{2})>0$
$\Rightarrow ({\cot }^{-1}x-3)({\tan }^{-1}x+2-\frac{\pi }{2})>0$
$\Rightarrow ({\cot }^{-1}x-3)(2-{\cot }^{-1}x)>0(\because {\tan }^{-1}x+{\cot }^{-1}x=\frac{\pi }{2})$
$\Rightarrow ({\cot }^{-1}x-3)({\cot }^{-1}x-2)<0$
$\Rightarrow 2<{\cot }^{-1}x<3$
$\Rightarrow \cot 3<x<\cot 2$
($\because {\cot }^{-1}x$ is a decreasing function)
Hence, $x\in (\cot 3,\cot 2)$