Solution :
We have, tan (A + B) = \(\sqrt{3}\)
\(\implies\) tan (A + B) = tan 60 \(\implies\) A + B = 60 …………(1)
Also, tan (A – B) = \(1\over \sqrt{3}\)
\(\implies\) tan(A – B) = tan 30 \(\implies\) A – B = 30 …………(2)
Solving (1) and (2), we get
A = 45 and B = 15