Given $$t_1=\tan x^{\tan x}$$
$$t_2=\tan x^{\cot x}$$
$$t_3=\cot x^{\tan x}$$
$$t_4=\cot x^{\cot x}$$
for $x\in(0, \frac{\pi}{4})$. Arrange them in descending order.
I tried in this way:
in $(0, \frac{\pi}{4})$ from the graphs of $\tan x$ and $\cot x$ $$\cot x \gt \tan x$$ and in $(0, \frac{\pi}{4})$ both are positive we have
$$\cot x^{\cot x} \gt \tan x^{\cot x}$$ that is
$$t_4 \gt t_2$$ and similarly
$$\cot x^{\tan x} \gt \tan x^{\tan x}$$ that is
$$t_3 \gt t_1$$
Now only comparison left is between $t_4$ and $t_1$.
if $f(x)=x^x$ it is decreasing in $(0, \frac{1}{e})$ and increasing from $(\frac{1}{e}, \frac{\pi}{4})$
so in $(0, \frac{1}{e})$ since $\cot x \gt \tan x$ we get
$$\cot x^{\cot x} \lt \tan x^{\tan x}$$ that is
$$t_4 \lt t_1$$
hence in $(0, \frac{1}{e})$ $$t_3 \gt t_1 \gt t_4 \gt t_2$$
and in $(\frac{1}{e}, \frac{\pi}{4})$ since $x^x$ is increasing we get $$\cot x^{\cot x} \gt \tan x^{\tan x}$$ that is
$$t_4 \gt t_1$$
but i cannot arrange them in decreasing order in $(\frac{1}{e}, \frac{\pi}{4})$
Best Answer
Set $y = \tan x$. Then $y \in (0,1)$ and
We got rid of the noise. Since $\log$ in increasing, this is the same of comparing
It's obvious that $$-\frac{1}{y} < -y < y < \frac{1}{y}$$ Multiplying by $\log y <0$ we get $$-\frac{1}{y}\log y > -y\log y > y \log y > \frac{1}{y} \log y$$
which means $$t_4>t_3>t_1>t_2$$