Inspired by a Physics Problem

Using half-angle tangent substitution and let $t = \tan \dfrac x2$ , we have $\sin x = \dfrac {2t}{1+t^2}$ , $\cos x = \dfrac {1-t^2}{1+t^2}$ , and $\tan x = \dfrac {2t}{1-t^2}$ .

Then $f(x) = \dfrac {1-\cos x}{\sin x} = \dfrac {1-\frac {1-t^2}{1+t^2}}{\frac {2t}{1+t^2}} = t = \tan \dfrac x2$ . For $0\le x \le \frac \pi 2$ , $\tan \dfrac x2$ is an increasing function. Therefore, $f \left(\frac \pi 3\right) > f \left(\frac \pi 4\right)$ .

And $g(x) = \dfrac {\color{#3D99F6}1-\cos x}{{\color{#3D99F6}\sin x}\tan x} = \dfrac {\color{#3D99F6}f(x)}{\tan x} = \dfrac t{\frac {2t}{1-t^2}} = \dfrac {1-t^2}2 = \dfrac {1-(f(x))^2}2$ $\implies g \left(\frac \pi 3\right) < g \left(\frac \pi 4\right)$ .

Therefore, the answer $\boxed{f \left(\frac \pi 3\right) > f \left(\frac \pi 4\right) \text{ and } g \left(\frac \pi 3\right) < g \left(\frac \pi 4\right)}$ .