Question : If $\tan (A+B)=\sqrt{3}$ and $\tan (A-B)=\frac{1}{\sqrt{3}}; 0°<(A+B)<90°; A > B$, then the values of $A$ and $B$ are respectively:
Option 1: 45° and 15°
Option 2: 15° and 45°
Option 3: 30° and 30°
Option 4: 60° and 30°
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Correct Answer: 45° and 15°
Solution :
Given: $\tan (A+B)=\sqrt{3}$ and $\tan (A-B)=\frac{1}{\sqrt{3}}$
$\tan (A+B)=\sqrt{3}=\tan60°$
⇒ $A+B=60°$ .........................(1)
Also, $\tan (A-B)=\frac{1}{\sqrt{3}}=\tan30°$
⇒ $A-B=30°$ ............................(2)
Solving equations (1) and (2) we get,
$A=45°$ and $B=15°$
Hence, the correct answer is 45° and 15°.
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