Motivated from Integral contest, a slight variation of it
$$\int_{0}^{\pi/2}{\ln\cos x\over \tan x}\cdot\ln\left({\ln\sin x\over \ln \cos x}\right)\mathrm dx={\pi^2\over 4!}\tag1$$
Making an attempt:
$$u={\ln\sin x\over \ln \cos x}\implies du={\cot x\ln\cos x+\tan x \ln\sin x\over \ln^2\cos x}dx\tag2$$
$$u=\ln\cos x\implies du=-\tan x dx\tag3$$
$$u=\ln\sin x\implies du=\cot x dx\tag4$$
Using $(4)$, then $(1)$ becomes
$$\int_{0}^{\infty}\ln\sqrt{1-e^{-2u}}\ln\left({\ln\sqrt{1-e^{-2u}}\over u}\right)\mathrm du\tag5$$
Recall $$\ln(1-x)=\sum_{n=1}^{\infty}{(-1)^n\over n}x^n\tag6$$
Then $(5)$ becomes
$$\sum_{n=1}^{\infty}{(-1)^n\over 2n}\color{red}{\int_{0}^{1}e^{-2un}\ln\left({\ln\sqrt{1-e^{-2u}}\over u}\right)\mathrm du}\tag7$$
The red part rewrite as
$$I_1-I_2={1\over 2}\int_{0}^{1}e^{-2un}\ln\left(\ln(1-e^{-2un})\right)\mathrm du-\int_{0}^{1}e^{-2un}\ln u\mathrm du\tag8$$
$$I_1-\color{blue}{I_2}={1\over 2}\int_{0}^{1}e^{-2un}\ln\left[\ln(1-e^{-2un})\right]\mathrm du-\color{blue}{{1\over 4}[Ei(-2n)-\gamma -\ln(2n)]}\tag9$$
Where Ei is the Exponential integral Ei
As for $I_1$ seems very difficult.
How can we prove $(1)?$