The value of $\tan\frac{5\pi}{6}$ is $-\frac{1}{\sqrt{3}}$, usually written in rationalised form as $-\frac{\sqrt{3}}{3} \approx -0.5774$.
Quick Answer:
Result: tan(5π/6) = −1/√3 = −√3/3 ≈ −0.5774
Notation: rationalised exact form −√3/3 (equivalently −1/√3)
Method shown: degree conversion + reference angle + sin/cos quotient
Degree equivalent: tan 150°
Sign: negative (second quadrant)
Quick Reference Table
Neighbouring angles in both notations, with their tangent values.
Angle (radians) | Angle (degrees) | Quadrant | $\tan$ value |
|---|---|---|---|
$\frac{\pi}{6}$ | 30° | I | $\frac{\sqrt{3}}{3}$ |
$\frac{\pi}{3}$ | 60° | I | $\sqrt{3}$ |
$\frac{2\pi}{3}$ | 120° | II | $-\sqrt{3}$ |
$\frac{5\pi}{6}$ | 150° | II | $-\frac{\sqrt{3}}{3}$ |
$\pi$ | 180° | — | $0$ |
$\frac{7\pi}{6}$ | 210° | III | $\frac{\sqrt{3}}{3}$ |
What Tangent of an Angle Means
Tangent is the ratio of sine to cosine: $\tan\theta = \frac{\sin\theta}{\cos\theta}$. On the unit circle it equals the $y$-coordinate divided by the $x$-coordinate of the terminal point, which is why tangent reads as the slope of the radius.
A quadrant is one of the four regions the axes divide the plane into, numbered I to IV anticlockwise. Tangent is positive where sine and cosine share a sign (quadrants I and III) and negative where they differ (II and IV). The angle 5π/6 is in the second quadrant, so its tangent is negative before any arithmetic. These sign patterns come straight from the reciprocal identities and the basic ratio definitions.
Methods to Find Tan 5pi/6
How do you find tan 5pi/6 without a calculator? Each method below lands on the same value.
Method 1: Convert radians to degrees
$$\frac{5\pi}{6} \times \frac{180°}{\pi} = \frac{5 \times 180°}{6} = 150°$$
So $\tan\frac{5\pi}{6} = \tan 150°$. Going the other way, 150° converts back by multiplying by $\frac{\pi}{180°}$. If the conversion factor feels shaky, the radian-to-degree relationship lays it out.
Final answer: $150°$.
Method 2: Reference angle
For a second-quadrant angle the reference angle is π minus the angle.
$$\pi - \frac{5\pi}{6} = \frac{6\pi - 5\pi}{6} = \frac{\pi}{6}$$
The reference angle is $\frac{\pi}{6}$ (30°), and $\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$.
Quadrant II makes tangent negative, so:
$$\tan\frac{5\pi}{6} = -\tan\frac{\pi}{6} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}$$
Final answer: $-\frac{\sqrt{3}}{3}$.
Method 3: Sine over cosine
At 150° the unit-circle point is $\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$. Tangent is $y$ over $x$:
$$\tan\frac{5\pi}{6} = \frac{\sin\frac{5\pi}{6}}{\cos\frac{5\pi}{6}} = \frac{\tfrac{1}{2}}{-\tfrac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}$$
Final answer: $-\frac{\sqrt{3}}{3}$.
Tan 5pi/6 is the radian twin of tan 150°; both describe the same 150° direction, so the two pages differ only in how the angle is written, not in the value.
Common Mistakes With Tan 5pi/6
Mistake 1: Leaving the answer as −1/√3 when rationalised form is expected
Where it slips in: At the final line, when $-\frac{1}{\sqrt{3}}$ looks finished.
Don't do this: Hand in $-\frac{1}{\sqrt{3}}$ on a paper that asks for a rationalised denominator.
The correct way: Multiply top and bottom by $\sqrt{3}$ to get $-\frac{\sqrt{3}}{3}$. Both are correct values; $-\frac{\sqrt{3}}{3}$ is the standard written form.
Mistake 2: Dropping the negative sign
Where it slips in: After computing the reference value $\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$, which is positive.
Don't do this: Report $\tan\frac{5\pi}{6} = \frac{\sqrt{3}}{3}$.
The correct way: The reference angle gives the magnitude; the second quadrant supplies a negative sign because sine and cosine have opposite signs there. The reciprocal $\frac{1}{\tan\theta}$ and the inverse $\tan^{-1}\theta$ are different ideas — don't let one stand in for the other when checking signs.
Mistake 3: Using 30° as the reference but adding instead of subtracting
Where it slips in: Confusing the second-quadrant rule with the third-quadrant one.
Don't do this: Compute $\frac{5\pi}{6} - \pi$ to get a reference angle.
The correct way: In quadrant II, reference angle = π − angle, so $\pi - \frac{5\pi}{6} = \frac{\pi}{6}$.
To build fluency with these second-quadrant values alongside a teacher, Bhanzu's trigonometry tutor and broader math tutoring walk through the reference-angle method on every quadrant.
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