Question

Using Lagrange's mean value theorem, prove that

(b − a) sec² a < tan b − tan a < (b − a) sec² b

where 0 < a < b < $\frac{\mathrm{\pi }}{2}$.

Answer 1

​Consider, the function
$f\left(x\right)=\tan x,x\in \left[a,b\right],0<a<b<\frac{\mathrm{\pi }}{2}$

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

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

Consequently, $c\in \left(a,b\right)$ such that $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$.

Now,
$f\left(x\right)=\tan x$ $\Rightarrow$ $f\text{'}\left(x\right)=se{c}^{2}x$, $f\left(a\right)=\tan a,f\left(b\right)=\tan b$

$\therefore$ $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$ $\Rightarrow$ $se{c}^{2}c=\frac{\tan b-\tan a}{b-a}...\left(1\right)$


Now,
$$\begin{gathered} c\in \left(a,b\right) \\ \Rightarrow a<c<b \\ \Rightarrow se{c}^{2}a<se{c}^{2}c<se{c}^{2}b\left[\because se{c}^{2}x\mathrm{is}\mathrm{increasing}\mathrm{in}\left(0,\frac{\mathrm{\pi }}{2}\right)\right] \\ \Rightarrow se{c}^{2}a<\frac{\tan b-\tan a}{b-a}<se{c}^{2}b\left[\mathrm{from}\left(1\right)\right] \\ \Rightarrow \left(b-a\right)se{c}^{2}a<\tan b-\tan a<\left(b-a\right)se{c}^{2}b \end{gathered}$$

Hence proved.