what is the exact value of tan(π/12) o 1/(2 + √3) o 1/√3 o 3/√3 o 1/(2 - √3)

Explanation:

Step1: Express $\frac{\pi}{12}$ as a difference

We know that $\frac{\pi}{12}=\frac{\pi}{3}-\frac{\pi}{4}$. And the formula for $\tan(A - B)=\frac{\tan A-\tan B}{1 + \tan A\tan B}$.

Step2: Find $\tan\frac{\pi}{3}$ and $\tan\frac{\pi}{4}$

We know that $\tan\frac{\pi}{3}=\sqrt{3}$ and $\tan\frac{\pi}{4}=1$.

Step3: Substitute values into the formula

$\tan(\frac{\pi}{3}-\frac{\pi}{4})=\frac{\tan\frac{\pi}{3}-\tan\frac{\pi}{4}}{1+\tan\frac{\pi}{3}\tan\frac{\pi}{4}}=\frac{\sqrt{3}- 1}{1+\sqrt{3}\times1}=\frac{\sqrt{3}-1}{1 + \sqrt{3}}$.

Step4: Rationalize the denominator

Multiply the numerator and denominator by $1-\sqrt{3}$: [ \begin{align*} \frac{\sqrt{3}-1}{1+\sqrt{3}}\times\frac{1 - \sqrt{3}}{1-\sqrt{3}}&=\frac{(\sqrt{3}-1)(1 - \sqrt{3})}{1-3}\ &=\frac{\sqrt{3}-3 - 1+\sqrt{3}}{-2}\ &=\frac{2\sqrt{3}-4}{-2}\ &=2-\sqrt{3}=\frac{1}{2 + \sqrt{3}} \end{align*} ]

Answer:

A. $\frac{1}{2+\sqrt{3}}$