${{\tan x}\over x}\gt {{x}\over {\sin x}}$ Problem was asked as an application of MVT theorem. I got the derivative of  $\sin x  \tan x - x^2$ as $\sin x +\tan x \sec x - 2x$. How do I prove that the derivative is positive for $x \in (0,\pi/2)$ so as to prove the function is increasing and prove the question.

Sorry I cannot find any relation about MVT and increasing/decreasing, but we may use monotonicity to deal with this problem.
Let $y>x$, for $x,y \in (0,\frac{\pi}{2})$
If $\sin{y}\tan{y}-y^2-\sin{x}\tan{x}+x^2>0$, function would be monotonically increasing.
$\because y-x>\sin({y-x})$, for $x,y \in (0,\frac{\pi}{2})$
$\therefore \sin{y}\tan{y}-y^2-\sin{x}\tan{x}+x^2$
$=\sin{y}\tan{y}-\sin{x}\tan{x}-(y^2-x^2)$
$=\sin{y}\tan{y}-\sin{x}\tan{x}-(y-x)(y+x)$
$>\sin{y}\tan{y}-\sin{x}\tan{x}-\sin({y-x})\sin({y+x})$
$=\frac{\sin^2{y}}{cos{y}}-\frac{\sin^2{x}}{cos{x}}+\frac{1}{2}(\cos{y}-\cos{x})$
$=\frac{(1-\cos^2{y})\cos{x}-(1-\cos^2{x})\cos{y}}{\cos{y}\cos{x}}+\frac{1}{2}(\cos{y}-\cos{x})$
$=\frac{(\cos{x}-\cos{y})(\cos{y}\cos{x}+1)}{\cos{x}\cos{y}}+\frac{1}{2}(\cos{y}-\cos{x})$
$=\frac{1}{2}\cos{x}\cos{y}+\frac{\cos{x}-\cos{y}}{\cos{y}\cos{x}}$
$\because \cos{x}\cos{y}>0$, for $x,y \in (0,\frac{\pi}{2})$
$\cos{x}-\cos{y}=-\frac{1}{2}\sin\frac{x+y}{2}\sin\frac{x-y}{2}=\frac{1}{2}\sin\frac{x+y}{2}\sin\frac{y-x}{2}>0$
$\therefore$ function is increasing on $(0,\frac{\pi}{2})$
