If $0<x<\frac{\pi}{2}$, then  

none of these

Consider the functions f(x) and g(x) given by

$f(x)=\tan x-x$ and $g(x)=x-\sin x, \text { for } 0<x<\frac{\pi}{2}$

We have,

$f'(x)=\sec ^2 x-1$ and $g'(x)=1-\cos x$

$\Rightarrow f'(x)>0$ and $g'(x)>0$ for all $x \in(0, \pi / 2)$

$\Rightarrow f(x)>f(0)$ and $g(x)>g(0)$ for all $x \in(0, \pi / 2)$

$\Rightarrow \tan x-x>0$ and $x-\sin x>0$ for all $x \in(0, \pi / 2)$

$\Rightarrow \tan x>x$ and $x>\sin x$ for all $x \in(0, \pi / 2)$

$\Rightarrow \sin x<x<\tan x$ for all $x \in(0, \pi / 2)$