If $f(x)=\frac{x}{\sin x}$ and $g(x)=\frac{x}{\tan x}$, where $0<x \leq 1$, then in this interval

f(x) is an increasing function

We have,

$f(x)=\frac{x}{\sin x}$ and $g(x)=\frac{x}{\tan x}$

∴  $f'(x)=\frac{\sin x-x \cos x}{\sin ^2 x}$ and $g'(x)=\frac{\tan x-x \sec ^2 x}{\tan ^2 x}$

Let $\phi(x)=\sin x-x \cos x$ and $\psi(x)=\tan x-x \sec ^2 x$

Then, $f(x)=\frac{\phi(x)}{\sin ^2 x}$ and $g'(x)=\frac{\psi(x)}{\tan ^2 x}$

Now,

$\phi'(x)=\cos x-\cos x+x \sin x=x \sin x$

and,

$\psi'(x)=\sec ^2 x-\sec ^2 x-2 x \sec ^2 x \tan x=-2 x \sec ^2 x \tan x$

For $0<x \leq 1$, we have

$x>0, \sin x>0, \tan x>0, \sec x>0$

$\Rightarrow \phi'(x)=x \sin x>0$ and $\psi'(x)<0$ for $0<x \leq 1$

$\Rightarrow \phi(x)$ is increasing on (0,1]  and $\psi(x)$ is decreasing on (0,1]

$\Rightarrow \phi(x)>\phi(0)$ and $\psi(x)<\psi(0)$

$\Rightarrow \phi(x)>0$ and $\psi(x)<0$

$\Rightarrow f'(x)=\frac{\phi(x)}{\sin ^2 x}>0$ and $g'(x)=\frac{\phi(x)}{\tan ^2 x}<0$

$\Rightarrow \quad f(x)$ is increasing on $(0,1]$ and $g(x)$ is decreasing on $(0,1]$.