Let $f(x)=\sin x-\tan x, x \in(0, \pi / 2)$ then tangent drawn to the curve $y=f(x)$ at any point will

lie above the curve

We have,

$y=\sin x-\tan x$

$\Rightarrow \frac{d y}{d x}=\cos x-\sec ^2 x$

$\Rightarrow \frac{d^2 y}{d x^2}=-\sin x-2 \sec ^2 x \tan x<0$ for all $x \in(0, \pi / 2)$

Hence, the tangent drawn to the curve will lie above the curve.