Question

If $f\left(x\right)=x\tan x,x\in (0,\frac{\pi }{2})$ then prove that $\frac{\tan x_2}{\tan x_1}>\frac{x_1}{x_2}$ where $0<x_1<x_2<\frac{\pi }{2}$.

Answer 1

$f\left(x\right)=x\tan x$

$f^{\prime }\left(x\right)=x{\sec }^{2}x+\tan x$

Now $\forall x\in (0,\frac{\pi }{2})$

$\Rightarrow {\sec }^{2}x>0,\tan x>o,x>c$

$\therefore f^{\prime }\left(x\right)>0$ $\forall x\in 0,\frac{\pi }{2}$

$f$ is increasing $\forall x\in [0,\frac{\pi }{2}]$

$\therefore f\left(x_2\right)>f\left(x_1\right)$ if $x_2>x_1$

$\therefore x_2\tan x_2>x_1\tan x_1$

$\therefore \frac{\tan x_2}{\tan x_4}>\frac{x_1}{x_2}$