Question

what is the exact value of \\( \tan \left( \frac { 19 \pi } { 12 } \
ight) \\)?
\\( \bigcirc - 2 - \sqrt { 3 } \\)
\\( \bigcirc - 2 + \sqrt { 3 } \\)
\\( \bigcirc 1 - \sqrt { 3 } \\)
\\( \bigcirc 1 + \sqrt { 3 } \\)

Explanation:

Step1: Rewrite the angle

We can rewrite \(\frac{19\pi}{12}\) as \(\frac{19\pi}{12}=2\pi-\frac{5\pi}{12}\). Then \(\tan(\frac{19\pi}{12})=\tan(2\pi - \frac{5\pi}{12})\).
Since \(\tan(A - B)=\frac{\tan A-\tan B}{1 + \tan A\tan B}\) and \(\tan(2\pi-\alpha)=-\tan\alpha\), so \(\tan(2\pi - \frac{5\pi}{12})=-\tan(\frac{5\pi}{12})\).
Also, \(\frac{5\pi}{12}=\frac{\pi}{4}+\frac{\pi}{6}\).

Step2: Use the tangent addition formula

The tangent addition formula is \(\tan(A + B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\).
Let \(A=\frac{\pi}{4}\) and \(B = \frac{\pi}{6}\), then \(\tan(\frac{\pi}{4}+\frac{\pi}{6})=\frac{\tan\frac{\pi}{4}+\tan\frac{\pi}{6}}{1-\tan\frac{\pi}{4}\tan\frac{\pi}{6}}\).
We know that \(\tan\frac{\pi}{4} = 1\) and \(\tan\frac{\pi}{6}=\frac{\sqrt{3}}{3}\).
Substitute these values into the formula: \(\tan(\frac{\pi}{4}+\frac{\pi}{6})=\frac{1+\frac{\sqrt{3}}{3}}{1 - 1\times\frac{\sqrt{3}}{3}}=\frac{\frac{3 + \sqrt{3}}{3}}{\frac{3-\sqrt{3}}{3}}=\frac{3+\sqrt{3}}{3-\sqrt{3}}\).
Rationalize the denominator: \(\frac{(3 + \sqrt{3})(3+\sqrt{3})}{(3-\sqrt{3})(3+\sqrt{3})}=\frac{9+6\sqrt{3}+3}{9 - 3}=\frac{12 + 6\sqrt{3}}{6}=2+\sqrt{3}\).

Step3: Find the value of \(\tan(\frac{19\pi}{12})\)

Since \(\tan(\frac{19\pi}{12})=-\tan(\frac{5\pi}{12})\) and \(\tan(\frac{5\pi}{12})=2+\sqrt{3}\), then \(\tan(\frac{19\pi}{12})=- (2+\sqrt{3})=-2-\sqrt{3}\).

Answer:

\(-2-\sqrt{3}\) (the first option)