Question

If the solution set of 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 $(a,b)$, then the value of ${\cot }^{-1}a+{\cot }^{-1}b$ is

Answer 1

$\left({\cot }^{-1}x\right)\left({\tan }^{-1}x\right)+(2-\frac{\pi }{2}){\cot }^{-1}x-3{\tan }^{-1}x-3(2-\frac{\pi }{2})>0$
Let ${\cot }^{-1}x=t$
$\Rightarrow t(\frac{\pi }{2}-t)+t(2-\frac{\pi }{2})-3(\frac{\pi }{2}-t)-3(2-\frac{\pi }{2})>0$
On simplifying we get,
$-t^2+5t-6>0$
$t^2-5t+6<0$
$2<t<3$
$2<co{t}^{-1}x<3$
$\cot 2>x>\cot 3$
Hence, $a=\cot 3,b=\cot 2$
${\cot }^{-1}a+{\cot }^{-1}b$
$={\cot }^{-1}\left(\cot 3\right)+{\cot }^{-1}\left(\cot 2\right)$
$=3+2$
$=5$