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Question:

Using the Mean Value Theorem, show that for all $0\lt a,b \lt \frac{\pi}{2}$ with $a\lt b$

$$\lvert \tan^{-1}(a)-\tan^{-1}(b)\rvert \lt \lvert a-b\rvert$$

My attempt:

Let $f(x)= \tan^{-1}(x)$

$\tan(x)\ \text{is continuous and differentiable on the interval so by the MVT,} $

$$\lvert \tan(b)-\tan(a) \rvert= \lvert b-a\rvert.f'(c)$$

$$f'(c)= \frac{1}{1+c^2}\lt \frac{1}{1+\frac{\pi^2}{4}}\lt1$$

Therefore we have,

$$\lvert \tan(a)-\tan(b) \rvert= \lvert a-b\rvert.f'(c)\lt1$$

Because $\lvert a-b\rvert= \lvert b -a\rvert$

Would this be correct?

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  • $\begingroup$ I'm sorry I don't quite see how, the reciprocal of tan is cot and the derivative of cot is -cosec$^2$(x) $\endgroup$ Commented May 24, 2016 at 0:57
  • $\begingroup$ $f'(c)$ need not be $\le 1$. $\endgroup$
    – Paul
    Commented May 24, 2016 at 0:57
  • $\begingroup$ What do you mean by $\tan^{-1}$, and why doesn't it appear in your attempt? (usually, $\tan^{-1}$ is the arctan function) $\endgroup$ Commented May 24, 2016 at 0:57
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    $\begingroup$ Did you mean $\frac{1}{\tan x}$ or $\arctan x$? These are very different things (and I hate that the notation $\tan^{-1}$ is used for this very reason). $\endgroup$ Commented May 24, 2016 at 0:58
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    $\begingroup$ When in doubt, replace $\operatorname{trig}^{-1}$ with $\operatorname{arctrig}$. It will make things so much less confusing for you. It is terrible notation that math teachers everywhere should be derided for for using and perpetuating. $\endgroup$ Commented May 24, 2016 at 1:00

1 Answer 1

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I have some sure that $\tan^{-1}a$ denotes that $\arctan a$ (otherwise, this doesn't hold, because you can let $a=b=\frac { pi}4$.

Therefore, $|\arctan a - \arctan b| = \frac{1}{1+\xi^2} |a-b| < |a-b|,$ where $\xi \in (a,b)$.

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  • $\begingroup$ Thanks Paul, is my edited version correct? $\endgroup$ Commented May 24, 2016 at 1:04
  • $\begingroup$ @THISISIT453 There exists some small mistakes still. $\endgroup$
    – Paul
    Commented May 24, 2016 at 4:09

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