What Does Tan 7pi/6 Mean?
Tangent is one of the three core trigonometric ratios: for an angle $\theta$, it is $\tan\theta = \frac{\sin\theta}{\cos\theta}$, and on a circle it reads as the $y$-coordinate divided by the $x$-coordinate of the point where the angle's radius lands. So $\tan\frac{7\pi}{6}$ asks a precise question about a specific spot on the circle.
The angle $\frac{7\pi}{6}$ measures $210^\circ$ once you convert from radians using $180^\circ = \pi$. That rotation stops in the third quadrant, the region past $\pi$ and before $\frac{3\pi}{2}$, where both coordinates are negative. Its reference angle, the acute gap to the horizontal axis, is $\frac{7\pi}{6} - \pi = \frac{\pi}{6}$.
Where Tan 7pi/6 Shows Up
A $210^\circ$ heading points into the third quadrant of a compass, and the slope of a straight path along that bearing carries a rise-over-run of $\tan\frac{7\pi}{6}$. The same value appears when a rotating machine part is tracked past the half-turn mark, where phase angles run beyond $\pi$ and the tangent describes the ratio of vertical to horizontal displacement.
Because tangent repeats every $\pi$ radians, $\tan\frac{7\pi}{6}$ equals $\tan\frac{\pi}{6}$, the slope of a gentle $30^\circ$ incline. That periodic reuse is exactly what makes tangent useful in applications of trigonometry that track repeating motion.
Standard-Angle Tangent Reference Table
Seven-sixths of $\pi$ is a rotation past the halfway mark of the circle, so its tangent reuses a first-quadrant value with a sign fixed by the quadrant. Here are the tangents you meet most, from $0$ around into the third quadrant.
Angle (degrees) | Angle (radians) | $\tan\theta$ (exact) | $\tan\theta$ (decimal) |
|---|---|---|---|
$0^\circ$ | $0$ | $0$ | $0.0000$ |
$30^\circ$ | $\dfrac{\pi}{6}$ | $\dfrac{1}{\sqrt{3}}$ | $0.5774$ |
$45^\circ$ | $\dfrac{\pi}{4}$ | $1$ | $1.0000$ |
$60^\circ$ | $\dfrac{\pi}{3}$ | $\sqrt{3}$ | $1.7321$ |
$90^\circ$ | $\dfrac{\pi}{2}$ | undefined | — |
$210^\circ$ | $\dfrac{7\pi}{6}$ | $\dfrac{1}{\sqrt{3}}$ | $0.5774$ |
Notice the last row repeats the second. That is not a coincidence: $210^\circ$ and $30^\circ$ share the same reference angle, so their tangents match in size, and both come out positive.
How Do You Find the Exact Value of Tan 7pi/6?
There are three clean routes, and all land on $\frac{1}{\sqrt{3}}$.
Method 1: The reference-angle rule.
Find the reference angle, take the first-quadrant tangent, then apply the quadrant sign.
$$\frac{7\pi}{6} - \pi = \frac{\pi}{6}, \qquad \tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$$
In the third quadrant tangent is positive (the ASTC rule, where the "T" quadrant keeps tangent positive), so:
$$\tan\frac{7\pi}{6} = +\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$$
Method 2: The unit circle.
Rotate the radius to $210^\circ$. The terminal point on the unit circle is $\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$.
$$\tan\frac{7\pi}{6} = \frac{y}{x} = \frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}$$
The two minus signs cancel, which is why a third-quadrant tangent is positive.
Method 3: Sine over cosine.
$$\sin\frac{7\pi}{6} = -\frac{1}{2}, \qquad \cos\frac{7\pi}{6} = -\frac{\sqrt{3}}{2}$$
$$\tan\frac{7\pi}{6} = \frac{\sin\frac{7\pi}{6}}{\cos\frac{7\pi}{6}} = \frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}$$
Examples of Tan 7pi/6
Working with $\frac{7\pi}{6}$ is really working with $\frac{\pi}{6}$ plus a sign rule. The set below builds from a plain evaluation to a rationalised form and an equation. One recurring pattern worth naming: students who have just learned that sine and cosine are negative in the third quadrant often expect the tangent to be negative too, and stamp a minus sign on an answer that should be positive.
Example 1
Evaluate $6\tan\frac{7\pi}{6}$.
$$6\tan\frac{7\pi}{6} = 6 \times \frac{1}{\sqrt{3}} = \frac{6}{\sqrt{3}} = 2\sqrt{3} \approx 3.464$$
Example 2
Find $\tan\frac{7\pi}{6}$ by deciding the sign first.
Wrong attempt. A student notes that $\frac{7\pi}{6}$ is past $\pi$, so sine and cosine are both negative, and concludes the tangent is negative: $-\frac{1}{\sqrt{3}}$.
That breaks under a quick check. Tangent is $\frac{\sin}{\cos}$, and here that is $\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}$. A negative divided by a negative is positive, so the minus sign cannot survive.
Correct. In the third quadrant, tangent is positive. The reference angle gives the size, $\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$, and the sign is $+$, so $\tan\frac{7\pi}{6} = \frac{1}{\sqrt{3}}$.
Example 3
Write $\tan\frac{7\pi}{6}$ in rationalised form.
$$\tan\frac{7\pi}{6} = \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}$$
Both $\frac{1}{\sqrt{3}}$ and $\frac{\sqrt{3}}{3}$ are correct; the rationalised $\frac{\sqrt{3}}{3}$ is the standard exam form.
Example 4
Confirm the periodic link $\tan\frac{7\pi}{6} = \tan\frac{\pi}{6}$.
Tangent repeats every $\pi$, so adding $\pi$ to an angle leaves the tangent unchanged.
$$\tan\left(\frac{\pi}{6} + \pi\right) = \tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$$
Since $\frac{\pi}{6} + \pi = \frac{7\pi}{6}$, the two are equal.
Example 5
Solve $\tan\theta = \frac{1}{\sqrt{3}}$ for $\theta$ in $[0, 2\pi)$.
Tangent hits $\frac{1}{\sqrt{3}}$ at its reference angle $\frac{\pi}{6}$, then again half a turn later.
$$\theta = \frac{\pi}{6} \quad \text{and} \quad \theta = \frac{\pi}{6} + \pi = \frac{7\pi}{6}$$
Both solutions are valid, which is why $\frac{7\pi}{6}$ turns up as a standard answer on the tangent.
Where Do Students Trip Up on Tan 7pi/6?
Mistake 1: Making the third-quadrant tangent negative
Where it slips in: Right after learning that sine and cosine are negative in the third quadrant, when the sign rule for tangent gets over-applied.
Don't do this: Writing $\tan\frac{7\pi}{6} = -\frac{1}{\sqrt{3}}$ because "everything is negative down there."
The correct way: Tangent is a ratio of two negatives in the third quadrant, so it is positive. The learner who anchors on ASTC instead of copying the sine sign gets this right every time; the one who reasons from a half-remembered rule is the one who flips it.
Mistake 2: Using the wrong reference angle
Where it slips in: Subtracting from the wrong axis, using $\frac{3\pi}{2} - \frac{7\pi}{6}$ or $2\pi - \frac{7\pi}{6}$ instead of measuring from $\pi$.
Don't do this: Taking the reference angle as $\frac{\pi}{3}$ and reporting $\tan\frac{7\pi}{6} = \sqrt{3}$.
The correct way: In the third quadrant, the reference angle is $\theta - \pi$. Here that is $\frac{7\pi}{6} - \pi = \frac{\pi}{6}$, giving $\frac{1}{\sqrt{3}}$, not $\sqrt{3}$.
Mistake 3: Confusing the value with tan 5pi/6
Where it slips in: Both angles have reference angle $\frac{\pi}{6}$, so the sizes match and only the sign separates them.
Don't do this: Reporting $\tan\frac{7\pi}{6}$ and $\tan\frac{5\pi}{6}$ as the same number.
The correct way: $\frac{5\pi}{6}$ is in the second quadrant where tangent is negative, so tan 5pi/6 $= -\frac{1}{\sqrt{3}}$, while $\tan\frac{7\pi}{6} = +\frac{1}{\sqrt{3}}$. Same size, opposite sign.
Key Takeaways
Tan 7π/6 equals $\frac{1}{\sqrt{3}}$, or $\frac{\sqrt{3}}{3}$, about $0.5774$, a positive value because the angle is in the third quadrant.
The reference angle is $\frac{\pi}{6}$, so the size matches $\tan\frac{\pi}{6}$, and periodicity gives $\tan\frac{7\pi}{6} = \tan\frac{\pi}{6}$.
The unit-circle point is $\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$, and $\frac{y}{x}$ returns the same $\frac{1}{\sqrt{3}}$.
The classic slip is reporting a negative answer; two negatives divide to a positive.
To work through more angles like this with a teacher, explore Bhanzu's trigonometry tutor sessions, a high school math tutor, or math classes online.
Practice These Before Moving On
Evaluate $\tan\frac{7\pi}{6} + \tan\frac{\pi}{6}$.
A bearing runs at $210^\circ$. Using $\tan\frac{7\pi}{6}$, find the rise for every $3$ units of horizontal run.
Solve $\sqrt{3},\tan\theta = 1$ for all $\theta$ in $[0, 2\pi)$.
Want a live Bhanzu trainer to walk through more tan 7pi/6 problems? Book a free demo class.
Read More
Sin, cos, and tan explained — the three core ratios and how they connect.
Tangent formula and derivation — where $\tan\theta = \frac{\sin\theta}{\cos\theta}$ comes from.
Trigonometric ratios of specific angles — the full standard-angle set.
Tan pi/6 exact value — the reference-angle twin that fixes the size of tan 7pi/6.
Tan 210 degrees — the same value written in degrees.
Coterminal angles — why angles a full turn apart share a tangent.
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