Question

The solution set of inequality $\left({\tan }^{-1}x\right)\left({\cot }^{-1}x\right)-\left({\tan }^{-1}x\right)(1+\frac{\pi }{2})-2{\cot }^{-1}x+2(1+\frac{\pi }{2})>\underset{y\to -\infty }{lim}[{\sec }^{-1}y-\frac{\pi }{2}]$ is
(where [.] denotes the greatest integer function)

• A. $(\tan 1,\tan 2)$
• B. $(-\cot 1,\cot 2)$
• C. $(-\tan 1,\tan 2)$
• D. $(-\tan 1,\infty )$

Answer 1

The correct option is D $(-\tan 1,\infty )$

$\because \underset{y\to -\infty }{lim}{\sec }^{-1}y={\frac{\pi }{2}}^{+}\Rightarrow \underset{y\to -\infty }{lim}[{\sec }^{-1}y-\frac{\pi }{2}]=0$

$\therefore \left({\tan }^{-1}x\right)\left({\cot }^{-1}x\right)-\left({\tan }^{-1}x\right)(1+\frac{\pi }{2})-2{\cot }^{-1}x+2(1+\frac{\pi }{2})>0$
$\Rightarrow ({\tan }^{-1}x-2)({\cot }^{-1}x-(1+\frac{\pi }{2}))>0$
$\Rightarrow ({\tan }^{-1}x+1)({\tan }^{-1}x-2)<0$
$\Rightarrow -1<{\tan }^{-1}x<2$
$\Rightarrow -1<{\tan }^{-1}x<\frac{\pi }{2}[\because {\tan }^{-1}\in (-\frac{\pi }{2},\frac{\pi }{2}\left)\right]\Rightarrow -\tan 1<x<\infty$