To find the correct ratio for \(\tan\left(\frac{7\pi}{6}\right)\), we begin by determining the coordinates of the angle \(\frac{7\pi}{6}\) in the unit circle.
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Finding the coordinates:
- The angle \(\frac{7\pi}{6}\) radians is in the third quadrant.
- The reference angle for \(\frac{7\pi}{6}\) is \(\frac{7\pi}{6} - \pi = \frac{7\pi}{6} - \frac{6\pi}{6} = \frac{\pi}{6}\).
- The coordinates for the angle \(\frac{\pi}{6}\) are \(\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\).
- In the third quadrant, both sine and cosine are negative, so the coordinates for \(\frac{7\pi}{6}\) are: \[ \left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right). \]
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Calculating \(\tan\left(\frac{7\pi}{6}\right)\):
- The tangent is given by: \[ \tan\left(\frac{7\pi}{6}\right) = \frac{\text{opposite}}{\text{adjacent}} = \frac{y}{x} = \frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}. \]
Now, let's see which option corresponds to this result.
A. Tan(7pi/6) = -sqrt3/2/-1/2 = sqrt 3: Incorrect
B. Tan(7pi/6)= -1/2/-sqrt3/w = sqrt3/3: Correct (if "w" stands for anything it does not affect the ratio)
C. Tan(7pi/6)=1/2/sqrt3/2 = sqrt 3/3: Incorrect, the simplification is incorrect
D. Tan(7pi/6)= -1/2/sqrt3/2 = -sqrt3/3: Incorrect
Thus, the correct answer is option B.