An equation!

Let $f(x) = \left(x^2 +1 \right) \ln x \; , \; x>0$ .

Prove that $f$ is strictly increasing on $(0, +\infty)$ .
If $x \in \left(0, \frac{\pi}{2} \right)$ , solve the equation
$\cos^2 x \cdot f \left ( \tan x \right ) + \sin^2 x \cdot f \left ( \cot^{22} x \right ) =0$

Solution

It is $\displaystyle f'(x) = x \left ( \ln x^2 + 1 + \frac{1}{x^2} \right )$ . However, recalling that
$\ln x \geq 1 - \frac{1}{x} \quad \text{foreach} \quad x>0$

we conclude that $f$ is strictly increasing.
We note that
$\begin{align*} \cos^2 x \cdot f \left ( \tan x \right ) + \sin^2 x \cdot f \left ( \cot^{22} x \right ) =0 &\Leftrightarrow \frac{\cos^2 x}{\sin^2 x} \cdot f \left ( \tan x \right ) + f \left ( \cot^{22}x \right ) = 0 \\ &\Leftrightarrow \cot^2 x \cdot f \left ( \frac{1}{\cot x} \right ) + f \left ( \cot^{22}x \right ) =0 \\ &\!\!\!\!\!\!\!\!\!\overset{u = \cot x}{\Leftarrow \! =\! =\! =\! =\! \Rightarrow } u^2 f \left ( \frac{1}{u} \right ) + f \left ( u^2 \right ) = 0 \end{align*}$

On the other hand, $f \left ( \frac{1}{x} \right ) = - \frac{f(x)}{x^2}$ and hence:

$\begin{align*} f \left ( u^{22} \right ) = f(u) &\Leftrightarrow u^{22} = u \\ &\Leftrightarrow u = 1 \\ &\Leftrightarrow x = \frac{\pi}{4} \end{align*}$