# 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