Question:
If $\sin (A+B)=1$ and $\tan (A-B)=\frac{1}{\sqrt{3}}, 0^{\circ}<(A+B) \leq 90^{\circ}$ and $A>B$ then find the values of $A$ and $B$.
Solution:
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.