What is $\tan(\alpha+2\beta)\tan(2\alpha+\beta)$, if $\sin(\alpha+\beta) = 1$ and $\sin(\alpha-\beta) = \frac 12$ If $\sin(\alpha+\beta) = 1$, $\sin(\alpha-\beta) = \frac 12$ , where $\alpha$, $\beta$ belongs to $[0,\frac \pi2],$ then what is $\tan(\alpha+2\beta)\tan(2\alpha+\beta)$?
What should be my approach to the question?

Since 
\begin{align}
0\leq \alpha \leq \frac{\pi}{2}, ~~~~~~~~~~~~~~0\leq \beta \leq \frac{\pi}{2},
\end{align}
we have
\begin{align}
0\leq \alpha+\beta \leq \pi,~~~~~~~~~~~~-\frac{\pi}{2}\leq \alpha-\beta \leq \frac{\pi}{2}.
\end{align}
Thus, from the equations given, we find that
\begin{align}
\sin(\alpha+\beta)&=1\Rightarrow \alpha+\beta=\frac{\pi}{2},\\
\sin(\alpha-\beta)&=1/2\Rightarrow \alpha-\beta=\frac{\pi}{6}.
\end{align}
Solving, we get $\alpha=\frac{\pi}{3}$ and $\beta=\frac{\pi}{6}$. Thus,
\begin{align}
\tan(2\alpha+\beta)\tan(\alpha+2\beta)=\left(-\sqrt{3}\right)\left(-\frac{1}{\sqrt{3}}\right)=1.
\end{align}
