Closed form of A Log-Sine-Cosine Integral by Mathematica

Question

By using the Mathematica 12.0, I found the numerical value of a Log-Sine-Cosine Integral that $$\int_0^{\frac{\pi}{6}} \log^2(2\sin\theta)\log(\tan(\theta/2))d\theta=-4.39530911884935704937076725879.$$ On the other hand, we have $$-\frac{40}{9}\beta(4)=-\frac{40}{9}\sum_{n=1}^\infty \frac{(-1)^{n-1}}{(2n-1)^4}=-4.39530911884935704937076725879.$$ Hence, the following equation seems to be correct $$\int_0^{\frac{\pi}{6}} \log^2(2\sin\theta)\log(\tan(\theta/2))d\theta=-\frac{40}{9}\beta(4) ?$$ But how to prove it?