If \( tan (A + B) \ = \ \sqrt{3} \) and \( tan (A – B) \ = \ \frac{1}{ \sqrt{3}} \)
\( 0° < \ A+B \ \le \ 90° \ ; \ A \ > \ B \) , find A and B.

\( tan (A + B) \ = \ \sqrt{3} \)
=> \( \ tan(A+B) \ = \ tan60° \)
\( \ A+B \ = \ 60°\) ............(1)

\( tan (A – B) \ = \ \frac{1}{ \sqrt{3}} \)
=> \( \ tan(A-B) \ = \ tan30° \)
\( A – B \ = \ 30° \)...............(2)

Solving (1) and (2), we get

\(\angle\)A = 45° and \(\angle\)B = 15°