If $\sin (\alpha+\beta)=1$ and $\sin (\alpha-\beta)=\frac{1}{2}$, where $0 \leq \alpha, \beta \leq \frac{\pi}{2}$, then find the values of $\tan (\alpha+2 \beta)$ and $\tan (2 \alpha+\beta)$.
Given:
$\sin (\alpha+\beta)=1$ and $\sin (\alpha-\beta)=\frac{1}{2}$
$\Rightarrow \alpha+\beta=90^{\circ} \quad \ldots(1)$
and $\alpha-\beta=30^{\circ} \quad \ldots(2)$
By adding eq $(1)$ and eq $(2)$ we get :
$2 \alpha=120^{\circ}$
$\Rightarrow \alpha=60^{\circ}$
By subtracting eq $(2)$ from eq $(1)$, we get:
$2 \beta=60^{\circ}$
$\Rightarrow \beta=30^{\circ}$
Therefore,
$\tan (\alpha+2 \beta)=\tan \left(60^{\circ}+2 \times 30^{\circ}\right)=\tan 120^{\circ}=-\sqrt{3}$
$\tan (2 \alpha+\beta)=\tan \left(2 \times 60^{\circ}+30^{\circ}\right)=\tan 150^{\circ}=-\frac{1}{\sqrt{3}}$