Using Lagrange's mean value theorem, prove that

​Consider, the function

$f(x)=\tan x, x \in[a, b], 0

Clearly, $f(x)$ is continuous on $[a, b]$ and derivable on $(a, b)$.

Thus, both the conditions of Lagrange's theorem are satisfied.

Consequently, $c \in(a, b)$ such that $f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$.

Now,

$f(x)=\tan x \Rightarrow f^{\prime}(x)=\sec ^{2} x, f(a)=\tan a, f(b)=\tan b$

$\therefore f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} \Rightarrow \sec ^{2} c=\frac{\tan b-\tan a}{b-a}$      ....(1)

Now, 

$c \in(a, b)$

$\Rightarrow a

$\Rightarrow \sec ^{2} a<\sec ^{2} c<\sec ^{2} b \quad\left[\because \sec ^{2} x\right.$ is increasing in $\left.\left(0, \frac{\pi}{2}\right)\right]$

$\Rightarrow \sec ^{2} a<\frac{\tan b-\tan a}{b-a}<\sec ^{2} b[$ from $(1)]$

$\Rightarrow(b-a) \sec ^{2} a<\tan b-\tan a<(b-a) \sec ^{2} b$

Hence proved.