please explain in detail-

Srujana

$cos(A-B)=\frac{1}{2}$

$\Rightarrow A-B =(2n \pi) \pm \frac{\pi}{3}$

$sin(A+B)=\frac{1}{2}$

$\Rightarrow A+B=(m \pi)+(-1)^m \frac{ \pi}{6}$

$\Rightarrow A=\left(n+\frac{m}{2})\pi \pm \frac{\pi}{6}+(-1)^m \frac{\pi}{12}$

$B=\left(-n+\frac{m}{2})\pi \mp \frac{\pi}{6}+(-1)^m \frac{\pi}{12}$

$m,n \in I$

$A-B =\frac{\pi}{3},\frac{5 \pi}{3},\frac{7 \pi}{3},...$

$A+B =\frac{\pi}{6},\frac{5 \pi}{6},\frac{7 \pi}{6},...$

for least positive values,

$A-B=\frac{\pi}{3} , A+B=\frac{5 \pi}{6}$

$\Rightarrow A=\frac{7 \pi}{12} ,B=\frac{\pi}{4}$