If $\sin(a+b)=1$ and $\sin(a-b)=\frac{1}{2}$ wher $a \geq 0$ and $b \leq \frac{\pi}{2}$ , find value of $\tan(a+2b)$ and $\tan(2a+b)$ If $\sin(a+b)=1$ and $\sin(a-b)=\frac{1}{2}$ where $a \geq 0$ and $b \leq \frac{\pi}{2}$ , find the value of $\tan(a+2b)$ and $\tan(2a+b)$.
I have tried to apply the formula for $\tan(a+(a+b))$ but couldn't reach anywhere.
THANKS

Hint:
$$\sin(a + b) = 1$$
$$\sin(a + b) = \sin \frac π2$$
$$a + b = \frac π2 = 90^\circ$$
Also,
$$\sin(a - b) = \frac 12$$
$$\sin(a - b) = \sin \frac π6$$
$$a - b = \frac π6 = 30^\circ$$
Solve equations to find $a$, $b$.
$$a = \frac π3 = 60^\circ$$
$$b = \frac π6 = 30^\circ$$
Put values of $a$ and $b$ to solve.
