In trigonometry, evaluate the exact value of tan(7π/6) using reference angles and quadrant signs.

Introduction / Context:
This trigonometric evaluation problem asks you to find the exact value of tan(7π/6). To solve it, you must recognize 7π/6 as a standard angle on the unit circle and use reference angles and quadrant information to determine the correct trigonometric value and sign. Such questions are typical in trigonometry courses and aptitude exams to test understanding of radian measure and unit circle properties.

Given Data / Assumptions:

• Angle: 7π/6 radians.
• We must find tan(7π/6) exactly.
• Standard special angles in radians, such as π/6, π/4, and π/3, and their tangent values are assumed known.

Concept / Approach:
First, express 7π/6 in terms of π to identify its position on the unit circle. Since 7π/6 = π + π/6, it lies in the third quadrant with reference angle π/6. In the third quadrant, both sine and cosine are negative, so tangent, as sin(θ)/cos(θ), is positive. Therefore, tan(7π/6) has the same magnitude as tan(π/6) but is positive. Using the known value of tan(π/6), we obtain the exact value of tan(7π/6).

Step-by-Step Solution:
1) Write 7π/6 as π + π/6. This shows that 7π/6 is π/6 beyond π. 2) When an angle is written as π + α, it lies in the third quadrant and has reference angle α. 3) Therefore, the reference angle for 7π/6 is π/6. 4) In the third quadrant, both sine and cosine are negative, but their ratio, tangent, is positive. 5) We know the special angle value tan(π/6) = 1 / √3. 6) Since tan(π + α) = tan(α) for any angle α, we get tan(7π/6) = tan(π + π/6) = tan(π/6) = 1 / √3.

Verification / Alternative check:
You can cross-check using the unit circle. The coordinates for angle π/6 are (√3 / 2, 1 / 2). For angle 7π/6 = π + π/6, both coordinates become negative: (−√3 / 2, −1 / 2). Tangent is given by y / x, so tan(7π/6) = (−1/2) / (−√3/2) = 1 / √3, confirming the earlier result. This verifies both the magnitude and the sign of the tangent value.

Why Other Options Are Wrong:
Option b (−1/√3) would be correct for a fourth quadrant angle with reference π/6, such as 11π/6, where tangent is negative. Option c (√3) and option d (−√3) correspond to reference angles of π/3 in different quadrants, not π/6. Option e (0) would be the tangent of angles like 0, π, or 2π, where the sine is zero. None of these match the situation for 7π/6, which clearly has a positive tangent of magnitude 1/√3.

Common Pitfalls:
A common error is misidentifying the quadrant for 7π/6 or confusing it with another angle such as 5π/6 or 11π/6. Another mistake is forgetting that tangent repeats every π radians, leading some students to compute tan(7π/6) incorrectly. Remembering that tan(π + α) = tan(α) and confirming the quadrant sign are key to avoiding such errors.

Final Answer:
The exact value of tan(7π/6) is 1/√3.