Please find exact value

**oceanmd** · January 5th 2008, 09:47 AM

Please find exact value
$<br /> tan \frac{19\pi}{12}$

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

**DivideBy0** · January 5th 2008, 09:57 AM

Originally Posted by oceanmd

Please find exact value
$<br /> tan \frac{19\pi}{12}$

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:

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

**oceanmd** · January 5th 2008, 07:01 PM

Thank you

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