If $0<a<b<\frac{\pi}{2}$ and $f(a, b)=\frac{\tan b-\tan a}{b-a}$, then

$f(a, b)>1$

Consider the function $f(x)=\tan x$ defined on $[a, b]$ such that $0<a<b<\frac{\pi}{2}$.

Applying Lagrange's mean value theorem, we have

$f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$ for some $c \in(a, b)$

$\Rightarrow\sec ^2 c=\frac{\tan b-\tan a}{b-a}$

$\Rightarrow f(a, b)=\sec ^2 c$

$\Rightarrow f(a, b)>1$           $\left[∵ \sec ^2 c>1 \text { as } c \in(0, \pi / 2)\right]$