Question

Anybody know the right answer?

Answer 1

Quadrant I : cot(x) > 0 → -cot(x) < 0

Quadrant II : cot(x) < 0 → -cot(x) > 0

Quadrant III : cot(x) > 0 → -cot(x) < 0

Quadrant IV : cot(x) < 0 → -cot(x) > 0

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$a)\ \left(\dfrac{11\pi}{6},\ -\sqrt3\right)\\\\\dfrac{11\pi}{6}\in\text{Quadrant IV, then -cot(x)} > 0,\ \text{we have}\ -\cot\dfrac{11\pi}{6}=-\sqrt3 < 0$

Answer: a)

$b)\\\dfrac{2\pi}{3}\in\text{Quadrant II},\ -\cot(x) > 0,\ \text{we have}\ \dfrac{\sqrt3}{3} > 0\\\\c)\\\dfrac{5\pi}{4}\in\text{Quadrant III},\ -\cot(x) < 0,\ \text{we have}\ -1 < 0\\\\d)\\\dfrac{4\pi}{3}\in\text{Quadrant III},\ -\cot(x) < 0,\ \text{we have}\ -\dfrac{\sqrt3}{3} < 0$

We know that one answer is only true, so we did not check the correctness of the cotangent value, only to the number sign