what is the exact value of \\( \\tan \\left( \\frac { 19 \\pi } { 12 } \\right) \\)?\n\\( \\bigcirc - 2…

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)