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Show that $\frac{ b - a} {1+b^2}< tan^{-1 }b - tan^{-1} a < \frac{b-a}{1+a^2}$ and hence deduce that $\frac{5\pi+4}{2} < tan^{-1}2 < \frac{\pi+2}{4}$

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  • $\begingroup$ This is difficult to read. Please see math.meta.stackexchange.com/questions/5020 $\endgroup$ Dec 19, 2017 at 6:47
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    $\begingroup$ Note that $\tan^{-1}b-\tan^{-1}a=\tan^{-1}\frac{b-a}{1+ab}$ $\endgroup$
    – QED
    Dec 19, 2017 at 6:48
  • $\begingroup$ @AbishankaSaha . So it's solvable by elementary trig. $\endgroup$ Dec 19, 2017 at 8:24

1 Answer 1

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use that $$\frac{\arctan(b)-\arctan(a)}{b-a}=\frac{1}{1+\xi^2}$$ and $$\xi \in (a,b)$$

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