Which of the following is equivalent to tan(5pi/6): tan(-pi/6), cot(5pi/6), tan(7pi/6), or tan(-5pi/6)?

Step-by-step solution

1. 1

   Locate the angle and get its exact value. $\frac{5\pi}{6} = 150^\circ$ sits in Quadrant II, where sine is positive and cosine negative, so tangent is negative. Its reference angle is $\pi-\frac{5\pi}{6} = \frac{\pi}{6}$, hence $\tan\frac{5\pi}{6} = -\tan\frac{\pi}{6} = -\frac{1}{\sqrt3} \approx -0.5774.$

2. 2

   Test tan(-pi/6) using the odd symmetry of tangent. $\tan(-\theta) = -\tan\theta$, so $\tan\!\left(-\frac{\pi}{6}\right) = -\tan\frac{\pi}{6} = -\frac{1}{\sqrt3}.$ This is an exact match.

3. 3

   Rule out cot(5pi/6). Cotangent is the reciprocal, not the same function: $\cot\frac{5\pi}{6} = \frac{1}{-1/\sqrt3} = -\sqrt3 \approx -1.7321$. Same sign but the wrong magnitude.

4. 4

   Rule out tan(7pi/6). Tangent has period $\pi$, so $\tan\frac{7\pi}{6} = \tan\!\left(\frac{7\pi}{6}-\pi\right) = \tan\frac{\pi}{6} = +\frac{1}{\sqrt3}$. The angle is in Quadrant III, where tangent is positive - wrong sign.

5. 5

   Rule out tan(-5pi/6). By oddness, $\tan\!\left(-\frac{5\pi}{6}\right) = -\tan\frac{5\pi}{6} = +\frac{1}{\sqrt3}$ - again the wrong sign, and in fact the negative of the target.

6. 6

   State the answer and the general rule. Only $\tan\!\left(-\frac{\pi}{6}\right)$ works. In general $\tan\theta = \tan(\theta+k\pi)$ for any integer $k$, and here $\frac{5\pi}{6}-\pi = -\frac{\pi}{6}$ - the two angles differ by exactly one period.