If tan (A + B) = √AAAA and tan (A – B) = (1 / √3)

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If tan (A + B) = √AAAA and tan (A – B) = (1 / √3) , ∠A + ∠B) < 90°, ∠A ≥ B, then ∠A is

1. 90°
2. 30°
3. 45°
4. 60°

Correct Option: C

tan(A + B) = √3 = tan60°
⇒ A + B = 60° ...(i)

tan(A – B) =	1	= tan30°
	√30

⇒ A – B = 30° ...(ii)
∴ A + B + A – B = 60° + 30°
⇒ 2A = 90°

⇒ A =	90°	= 45°
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