The solution set of the inequality (tan^{-1} | vedclass

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(C) Let $u = 	an^{-1} x$. Then $\cot^{-1} x = \frac{\pi}{2} - u$.Given inequality: $u(\frac{\pi}{2} - u) - u(1 + \frac{\pi}{2}) - 2(\frac{\pi}{2} - u

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$(-	an 1, 	an 2)$

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(C) Let $u = an^{-1} x$. Then $\cot^{-1} x = \frac{\pi}{2} - u$.Given inequality: $u(\frac{\pi}{2} - u) - u(1 + \frac{\pi}{2}) - 2(\frac{\pi}{2} - u) + 2(1 + \frac{\pi}{2}) > \lim_{x o \infty} [\sec^{-1} x - \frac{\pi}{2}]$.As $x o \infty$,$\sec^{-1} x o \frac{\pi}{2}$,so the limit is $[\frac{\pi}{2} - \frac{\pi}{2}] = [0] = 0$.Simplifying the left side: $\frac{\pi}{2}u - u^2 - u - \frac{\pi}{2}u - \pi + 2u + 2 + \pi > 0$.$-u^2 + u + 2 > 0 \Rightarrow u^2 - u - 2 $(u - 2)(u + 1) Since $u = an^{-1} x$,we have $-1 Taking tangent on all sides: $ an(-1) Thus,$- an 1 < x < an 2$.

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

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