Question

which of the following best explains why \\(\tan\frac{5\pi}{6} \
eq \tan\frac{5\pi}{3}\\)?
\\(\circ\\) the angles do not have the same reference angle.
\\(\circ\\) tangent is positive in the second quadrant and negative in the fourth quadrant.
\\(\circ\\) tangent is negative in the second quadrant and positive in the fourth quadrant.
\\(\circ\\) the angles do not have the same reference angle or the same sign.

Explanation:

Step 1: Find reference angles

For \(\theta = \frac{5\pi}{6}\), it's in the second quadrant. The reference angle is \(\pi - \frac{5\pi}{6}=\frac{\pi}{6}\). For \(\theta=\frac{5\pi}{3}\), it's in the fourth quadrant. The reference angle is \(2\pi - \frac{5\pi}{3}=\frac{\pi}{3}\). So reference angles are different.

Step 2: Determine sign of tangent

In the second quadrant, \(x\) is negative and \(y\) is positive, so \(\tan\theta=\frac{y}{x}\) is negative. In the fourth quadrant, \(x\) is positive and \(y\) is negative, so \(\tan\theta=\frac{y}{x}\) is negative? Wait, no, wait: Wait, \(\frac{5\pi}{3}\) is in fourth quadrant, \(x>0,y<0\), so \(\tan\frac{5\pi}{3}=\frac{y}{x}<0\)? Wait, no, I made a mistake. Wait \(\frac{5\pi}{6}\): second quadrant, \(x = r\cos\theta\), \(\cos\frac{5\pi}{6}=-\frac{\sqrt{3}}{2}\), \(y = r\sin\frac{5\pi}{6}=\frac{1}{2}\), so \(\tan\frac{5\pi}{6}=\frac{y}{x}=\frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}}=-\frac{1}{\sqrt{3}}\) (negative). \(\frac{5\pi}{3}\): fourth quadrant, \(\cos\frac{5\pi}{3}=\frac{1}{2}\), \(\sin\frac{5\pi}{3}=-\frac{\sqrt{3}}{2}\), so \(\tan\frac{5\pi}{3}=\frac{y}{x}=\frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=-\sqrt{3}\) (negative). Wait, but the options: Wait, maybe I messed up the quadrant signs. Wait, tangent is positive in first and third, negative in second and fourth. So both second and fourth quadrants have negative tangent? But the options say "tangent is negative in second and positive in fourth" which is wrong, "positive in second and negative in fourth" is wrong. Wait, no, let's re - evaluate the reference angles and signs.

Wait, the reference angle for \(\frac{5\pi}{6}\) is \(\frac{\pi}{6}\), for \(\frac{5\pi}{3}\) is \(\frac{\pi}{3}\). So reference angles are different. Now, sign: in second quadrant, tangent is negative (since \(\sin\) positive, \(\cos\) negative, \(\tan=\frac{\sin}{\cos}\) negative). In fourth quadrant, \(\sin\) negative, \(\cos\) positive, so \(\tan=\frac{\sin}{\cos}\) negative. Wait, but the options: option C says "tangent is negative in second and positive in fourth" which is incorrect. Wait, maybe I made a mistake in the angle \(\frac{5\pi}{3}\). Wait \(\frac{5\pi}{3}\) is \(300^{\circ}\), which is in fourth quadrant, \(\tan300^{\circ}=\tan(360 - 60)=-\tan60^{\circ}=-\sqrt{3}\) (negative). \(\frac{5\pi}{6}\) is \(150^{\circ}\), \(\tan150^{\circ}=\tan(180 - 30)=-\tan30^{\circ}=-\frac{1}{\sqrt{3}}\) (negative). Wait, but the options: let's check the options again.

Option A: "The angles do not have the same reference angle." Reference angle of \(\frac{5\pi}{6}\) is \(\frac{\pi}{6}\), reference angle of \(\frac{5\pi}{3}\) is \(\frac{\pi}{3}\), so they are different. But what about the sign? Wait, both tangents are negative, so sign is same? But the value of \(\tan\frac{5\pi}{6}=-\frac{1}{\sqrt{3}}\), \(\tan\frac{5\pi}{3}=-\sqrt{3}\), which are not equal. The reason is that they have different reference angles (so the magnitude from reference angle is different) and wait, but sign is same? Wait, no, maybe I made a mistake. Wait, no, the reference angle for \(\frac{5\pi}{6}\) is \(\frac{\pi}{6}\), so \(\tan\) of reference angle is \(\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}}\), and since it's in second quadrant, \(\tan\frac{5\pi}{6}=-\frac{1}{\sqrt{3}}\). For \(\frac{5\pi}{3}\), reference angle is \(\frac{\pi}{3}\), \(\tan\frac{\pi}{3}=\sqrt{3}\), and since it's in fourth quadrant, \(\tan\frac{5\pi}{3}=-\sqrt{3}\). So the reference angles are different (\(\frac{\pi}{6}\) vs \(\frac{\pi}{3}\)), so the "magnitude" from the reference angle is different, and even though both are negative, the values are different because reference angles are different. But the options:

Wait, option D: "The angles do not have the same reference angle or the same sign." Wait, but sign is same (both negative). Wait, maybe I messed up the sign. Wait, no, let's recalculate \(\tan\frac{5\pi}{3}\). \(\sin\frac{5\pi}{3}=-\frac{\sqrt{3}}{2}\), \(\cos\frac{5\pi}{3}=\frac{1}{2}\), so \(\tan\frac{5\pi}{3}=\frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=-\sqrt{3}\) (negative). \(\tan\frac{5\pi}{6}=\frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}}=-\frac{1}{\sqrt{3}}\) (negative). So sign is same. But reference angles are different (\(\frac{\pi}{6}\) and \(\frac{\pi}{3}\)). Wait, but the options:

Wait, maybe the question is why \(\tan\frac{5\pi}{6} eq\tan\frac{5\pi}{3}\). Let's compute the values:

\(\tan\frac{5\pi}{6}=\tan(150^{\circ})=-\frac{1}{\sqrt{3}}\)

\(\tan\frac{5\pi}{3}=\tan(300^{\circ})=-\sqrt{3}\)

They are not equal. The reference angles are different (\(\frac{\pi}{6}\) vs \(\frac{\pi}{3}\)), so the "base" tangent values from reference angles are different, and even though both are negative, the values are different. Now let's check the options:

Option A: "The angles do not have the same reference angle." Reference angle of \(\frac{5\pi}{6}\) is \(\pi-\frac{5\pi}{6}=\frac{\pi}{6}\), reference angle of \(\frac{5\pi}{3}\) is \(2\pi - \frac{5\pi}{3}=\frac{\pi}{3}\). So they don't have the same reference angle. But what about the sign? Wait, both are negative, so sign is same. But the option A says that's the reason. But wait, let's check other options.

Option B: "Tangent is positive in the second quadrant and negative in the fourth quadrant." Tangent is negative in second (since \(\sin\) positive, \(\cos\) negative, \(\tan=\frac{\sin}{\cos}\) negative) and negative in fourth ( \(\sin\) negative, \(\cos\) positive, \(\tan=\frac{\sin}{\cos}\) negative). So B is wrong.

Option C: "Tangent is negative in the second quadrant and positive in the fourth quadrant." Tangent is negative in fourth (as we saw), so C is wrong.

Option D: "The angles do not have the same reference angle or the same sign." Sign is same (both negative), so D is wrong.

Wait, but according to our calculation, reference angles are different, and that's a reason. So the correct option is A? Wait, but when we calculated, sign is same, but reference angles are different. So the reason why they are not equal is that they have different reference angles. Because if two angles have the same reference angle and same sign (same quadrant or opposite quadrants with same sign), then their tangents would be equal (since \(\tan\theta=\pm\tan(\text{reference angle})\)). Here, reference angles are different, so even though sign is same, the \(\tan(\text{reference angle})\) values are different (\(\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}}\), \(\tan\frac{\pi}{3}=\sqrt{3}\)), so the tangents of the original angles are different (both negative, but different magnitudes). So the main reason is that they don't have the same reference angle.

Answer:

A. The angles do not have the same reference angle.