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?
 A: Where does $\sin{x} = 1?$ In this case, $x = \pi/2.$  Where does $\sin{x} = 1/2?$ At $x = \pi/6$ in this case.  This gives you your system,
$$ \alpha+ \beta = \pi/2; \;\;\; \alpha - \beta = \pi/6.$$
From here, one can solve for $\alpha$ and $\beta$ and then plug in to get the answer.
A: Well, knowing what angles in the domain $[0,\frac{\pi}{2}]$ give you $\sin(\cdot)=1$ and $\sin(\cdot)=\frac{1}{2}$, we can say that 
$$
\alpha + \beta = \pi/2
$$
and 
$$
\alpha - \beta = \pi/6
$$
This is a system of two equations with two unknowns.  After you know $\alpha$ and $\beta$, you can plug in.
A: 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}
A: Directly calculate values of $\alpha$ and $\beta$.
Since the angles lie in $[0; \dfrac{\pi}2]$, we get $\alpha+\beta=\dfrac{\pi}2$ and $\alpha-\beta=\dfrac{\pi}6$. Solve the system of linear equations and substitute into product of tangents. Can you proceed it from here?
A: Just for the sake of a different approach - 
We can make an observation first.
If sin($\alpha$+$\beta$) = 1, then cos($\alpha$+$\beta$)=0; no matter what values $\alpha$ and $\beta$ take. Now, let me replace "$\alpha$+$\beta$" by another variable say "$A$". So, we have cos$A$ = 0. Therefore - 
cos$3A$ = 4cos$^3A$ - 3cos$A$ = 4(0) - 3(0) = 0.
Thus, cos$3A$ = 0 = cos($3\alpha$+$3\beta$) ............ (1)
Now, writing our expression - 
tan($\alpha$+$2\beta$)tan($2\alpha$+$\beta$) 
= $\frac{2sin(\alpha+2\beta)sin(2\alpha+\beta)}{2cos(\alpha+2\beta)cos(2\alpha+\beta)}$
=$\frac{cos(\alpha-\beta) - cos(3\alpha+3\beta)}{cos(3\alpha+3\beta) + cos(\alpha-\beta)}$
=$\frac{cos(\alpha-\beta) - 0}{0 + cos(\alpha-\beta)}$   (from equation 1)
= 1
