If $\sin(\alpha+\beta)=1,\ \sin(\alpha-\beta)=\frac{1}{2},\ \alpha,\beta\in\left[0,\frac{\pi}{2}\right]$ then the value of $\tan(\alpha+2\beta)\tan(2\alpha+\beta)$ is

$ \sin(\alpha+\beta)=1 $ $ \Rightarrow \alpha+\beta=\frac{\pi}{2} $ $ \sin(\alpha-\beta)=\frac{1}{2} $ $ \Rightarrow \alpha-\beta=\frac{\pi}{6} $ Solving, $ \alpha=\frac{\pi}{3},\ \beta=\frac{\pi}{6} $ $ \tan(\alpha+2\beta)\tan(2\alpha+\beta) $ $ =\tan\left(\frac{\pi}{3}+\frac{\pi}{3}\right)\tan\left(\frac{2\pi}{3}+\frac{\pi}{6}\right) $ $ =\tan\left(\frac{2\pi}{3}\right)\tan\left(\frac{5\pi}{6}\right) $ $ =(-\sqrt{3})\left(-\frac{1}{\sqrt{3}}\right)=1 $