If sin (A + B) = 1 and tan (A−B)

As we know that,

$\sin 90^{\circ}=1$

Thus,

if $\sin (A+B)=1$

$\Rightarrow A+B=90^{\circ} \quad \ldots(1)$

and $\tan 30^{\circ}=\frac{1}{\sqrt{3}}$

Thus,

if $\tan (A-B)=\frac{1}{\sqrt{3}}$

$\Rightarrow A-B=30^{\circ} \quad \ldots(2)$

Solving (1) and (2), we get

$A=60^{\circ}$ and $B=30^{\circ}$

Hence, the values of $A$ and $B$ are $60^{\circ}$ and $30^{\circ}$, respectively.