Prove that if $0<a<b<\pi/2$ then $\tan^{-1}b-\tan^{-1} a<\tan b -\tan a$

Question

Prove that if $0<a<b<\pi/2$ then: $$\tan^{-1}b-\tan^{-1} a<\tan b -\tan a$$

I have managed to prove that for each $x\in (0,\pi/2)$, $\tan x>\tan^{-1}x$, but I'm not sure this is the right direction

Answer 1

using the MVT namely $$\frac{f(b)-f(a)}{b-a}=f'(\xi)$$ and we get $$\frac{\arctan(x)-\arctan(y)}{x-y}=\frac{1}{1+\xi^2}$$

Answer 2

When solving i assume that $\tan ^{-1} = {1\over \tan}$ and not $\arctan $. Luckly the problem still holds.

Since $0<a<b<\pi/2$ we have $\tan (a-b)<0$ so

$$ {\tan a-\tan b \over 1+\tan a\tan b }<0$$ so

$$(\tan a-\tan b) \cdot (1+\tan a\tan b )<0$$

so $$\tan a-\tan b < \tan a\tan b (\tan b-\tan a)\;\;\;\;/:\tan a\tan b$$ and we get (since $0<a<b<\pi/2$ we have $\tan a>0$ and $\tan b>0$)

$$ \tan^{-1}b-\tan^{-1} a<\tan b -\tan a$$

Answer 3

A direct corollary of the Mean value theorem asserts that

Let $I\,$ be an interval, $x_0\in I$. If two functions $f$ and $g$, defined on $I$, satisfy $$\begin{cases} f(x_0)\le g(x_0),\\[1.5ex] f'(x)<g'(x) \enspace\forall x>x_0,\:x\in I, \end{cases}$$ then $\;f(x)<g(x)$ for all $x>x_0$, $x\in I$.