# Math Help - Please find exact value

1. ## Please find exact value

$
tan \frac{19\pi}{12}$

The correct answer is $-2-{\sqrt 3}$

$tan \frac{19\pi}{12}$

-tan 75=-tan (45 +30)=- $\frac{3+\sqrt 3}{3-\sqrt 3}$

What am I doing wrong?

Thank you

2. Originally Posted by oceanmd
$
tan \frac{19\pi}{12}$

The correct answer is $-2-{\sqrt 3}$

$tan \frac{19\pi}{12}$

-tan 75=-tan (45 +30)=- $\frac{3+\sqrt 3}{3-\sqrt 3}$

What am I doing wrong?

Thank you
You are correct, you just need to rationalize the denominator:

$-\frac{3+\sqrt{3}}{3-\sqrt{3}}=-\frac{3+\sqrt{3}}{3-\sqrt{3}}\left(\frac{3+\sqrt{3}}{3+\sqrt{3}}\right )=-\frac{12+6\sqrt{3}}{6}=-(2+\sqrt{3})=-2-\sqrt{3}$

3. Thank you