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

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