If cos x = - $\frac{5}{13}$ where 180° < X < 270°, what is the value of tan x -sin x ?

Analyze the question: We're given cos x = -5/13 with x in the third quadrant (180° < x < 270°). We need tan x - sin x.

Concept: In the third quadrant, sine and cosine are both negative, and tangent is positive. Use the 5-12-13 right triangle (or sin²x + cos²x = 1) to find sin x.

Solution:

1. Find sin x: sin²x = 1 - cos²x = 1 - 25/169 = 144/169 sin x = ±12/13, and since x is in the third quadrant, sin x = -12/13.

2. Find tan x: tan x = sin x / cos x = (-12/13) / (-5/13) = 12/5

3. Compute tan x - sin x: tan x - sin x = 12/5 - (-12/13) = 12/5 + 12/13 = (12×13 + 12×5) / 65 = (156 + 60) / 65 = 216/65

Answer: 216/65, which is option C.

Tip: Remember "ASTC" (All, Sine, Tangent, Cosine) for the signs of trig ratios in each quadrant. In the third quadrant, only tangent is positive. Also, watch the double negative: subtracting a negative sine turns into addition.