the exact value of tan 13π/12 is (simplify your answer. use integers or fractions for any numbers in the…

Explanation:

Step1: Rewrite the angle

We can write $\frac{13\pi}{12}=\frac{3\pi}{4}+\frac{\pi}{3}$.

Step2: Use the tangent - addition formula

The tangent - addition formula is $\tan(A + B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$. Here $A=\frac{3\pi}{4}$ and $B = \frac{\pi}{3}$. We know that $\tan\frac{3\pi}{4}=- 1$ and $\tan\frac{\pi}{3}=\sqrt{3}$.

Step3: Substitute the values

Substitute $\tan A=-1$ and $\tan B = \sqrt{3}$ into the formula: [ \begin{align*} \tan\left(\frac{3\pi}{4}+\frac{\pi}{3}\right)&=\frac{\tan\frac{3\pi}{4}+\tan\frac{\pi}{3}}{1 - \tan\frac{3\pi}{4}\tan\frac{\pi}{3}}\ &=\frac{-1+\sqrt{3}}{1-(-1)\times\sqrt{3}}\ &=\frac{-1 + \sqrt{3}}{1+\sqrt{3}} \end{align*} ]

Step4: Rationalize the denominator

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

Answer:

$2-\sqrt{3}$