If $0<\alpha<\beta<\frac{\pi}{2}$, then

$\frac{\tan \beta}{\tan \alpha}>\frac{\alpha}{\beta}$ 

Consider the function f(x) given by

$f(x)=x \tan x, x \in(0, \pi / 2)$

We have,

$f'(x)=x \sec ^2 x+\tan x>0$ for all $x \in(0, \pi / 2)$

⇒ f(x) is increasing on $(0, \pi / 2)$

$\Rightarrow f(\alpha)<f(\beta)$ for $0<\alpha<\beta<\frac{\pi}{2}$

$\Rightarrow \alpha \tan \alpha<\beta \tan \beta$

$\Rightarrow \frac{\alpha}{\beta}<\frac{\tan \beta}{\tan \alpha}$