# Using Mean Value Theorem to prove an inequality

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?

• I'm sorry I don't quite see how, the reciprocal of tan is cot and the derivative of cot is -cosec$^2$(x) Commented May 24, 2016 at 0:57
• $f'(c)$ need not be $\le 1$.
– Paul
Commented May 24, 2016 at 0:57
• What do you mean by $\tan^{-1}$, and why doesn't it appear in your attempt? (usually, $\tan^{-1}$ is the arctan function) Commented May 24, 2016 at 0:57
• 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). Commented May 24, 2016 at 0:58
• 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. Commented May 24, 2016 at 1:00

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 $\xi \in (a,b)$.