If sin (α + β) = 1 and sin (α − β) =

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,

$\tan (\alpha+2 \beta)=\tan \left(60^{\circ}+2 \times 30^{\circ}\right)=\tan 120^{\circ}=-\sqrt{3}$

$\tan (2 \alpha+\beta)=\tan \left(2 \times 60^{\circ}+30^{\circ}\right)=\tan 150^{\circ}=-\frac{1}{\sqrt{3}}$