The function $f(x)=\frac{x}{1+x \tan x}$ has

one point of maximum in the interval $(0, \pi / 2)$

We have,

$f(x) =\frac{x}{1+x \tan x}$

$\Rightarrow f'(x) =\frac{1+x \tan x-x\left(\tan x+x \sec ^2 x\right)}{(1+x \tan x)^2}$

$\Rightarrow f'(x) =\frac{1-x^2 \sec ^2 x}{(1+x \tan x)^2}$

For local maximum or minimum, we must have

$f'(x)=0$

$\Rightarrow \frac{1-x^2 \sec ^2 x}{(1+x \tan x)^2}=0 \Rightarrow \cos ^2 x=x^2 \Rightarrow \cos x= \pm x$

$\Rightarrow \cos x=x$       [∵ x and cos x are positive for $x \in(0, \pi / 2)$]

In order to find the number of solutions of this equation, let us draw the graphs of the curves $y=\cos x$ and $y=x$ in $(0, \pi / 2$ ). We find that the two curves intersect at one point $(\alpha, \alpha)$ only.

∴  $f'(x)=0 \Rightarrow \cos \alpha=\alpha$

Now,

$f'(x)=\frac{1-x^2 \sec ^2 x}{(1+x \tan x)^2}$

$\Rightarrow (1+x \tan x)^2 f'(x)=1-x^2 \sec ^2 x$

$\Rightarrow (1+x \tan x)^2 f''(x)+2(1+x \tan x)\left(x \sec ^2 x+\tan x\right) f'(x) \\
=-2 x \sec ^2 x-2 x^2 \sec ^2 x \tan x$

$\Rightarrow (1+\alpha \tan \alpha)^2 f''(\alpha)=-2 \alpha \sec ^2 \alpha-2 \alpha^2 \sec ^2 \alpha \tan \alpha$

$\Rightarrow f''(\alpha)<0$

Therefore, $x=\alpha$ is a point of local maximum

Hence, $f(x)$ attains local maximum at one-point in the interval $(0, \pi / 2)$.