Question : Using $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}$, find the value of $\tan 15°$.
Option 1: $\sqrt{3}+1$
Option 2: $\sqrt{3}-1$
Option 3: $2-\sqrt{3}$
Option 4: $2+\sqrt{3}$
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Correct Answer: $2-\sqrt{3}$
Solution :
Given: $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}$
⇒ $\tan (45°-30°)=\frac{\tan 45°-\tan 30°}{1+\tan 45° \tan 30°}$
⇒ $\tan15° = \frac{(1-\frac{1}{\sqrt3})}{(1+\frac{1}{\sqrt3})}$
⇒ $\tan15° = \frac{(\sqrt3-1)}{(\sqrt3+1)}$
After rationalizing,
⇒ $\tan15°=2-\sqrt{3}$
Hence, the correct answer is $2-\sqrt{3}$.
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