Given that sec θ = -2 and θ lies in the third quadrant, calculate the value of cot θ.

Answer 1

sec θ = -2
cos θ = -1/2

This is where I like to think about the unit circle.
Remember, on the unit circle, the hypotenuse is 1, so the x-coordinate is the cos value. (the angle originates from the center, so the 'adjacent' side would be on the x-axis. cos = adjacent/hypotenuse and b/c hypotenuse is 1, the cos is just the x-axis.)

So, when the x-value is -1/2 on the unit circle, it has to be on the 'left' side, so the second or third quadrant, and the problem lets us know that it is in the third quadrant. So what reference angle gives a cos value of 1/2? Well, the way I like to think of it is that there are several possibilities of cos and sin values. Either it's 1/2, sqrt2/2, or sqrt3/2. Obviously, there are negative values, too.
For this problem the cos is -1/2 and so the sin must be sqrt3/2. (I think of 1/2 and sqrt3/2 as pairs, with sqrt2/2 applying only angles with a radian value with a denominator of 4) We know that the length of the cos, 1/2, is smaller in length than the height of the sin, sqrt3/2. So if we think of a triangle (because that's where the angle comes from) that has the length of cos and height of sin, our triangle in this case must have a larger central angle than if the sin was 1/2 and the cos was root3/2. If you have trouble with this concept, I would just draw it out! Instead of memorizing the unit circle just draw the triangle and see if the angle would be bigger or smaller. If it is bigger, then it is pi/3 (60 degrees) and if it is smaller, then it is pi/6. So, in our case, because the sin value is larger than the cos value, our reference angle must be bigger, so, pi/3.

Because our angle is in the third quadrant, it protrudes that pi/3 angle from the negative side on the x-axis, so really it is pi + pi/3, which is 4pi/3. But really, we did not even need to know what the angle was. From the cos we already figured out the sin. cos = -1/2 sin = sqrt3/2
cot = cos/sin
cot=-1/sqrt3 = -sqrt3/3

Sorry if this was confusing.

Answer 2

recall that on the circle, at angle θ,
sinx = y/r
cosx = x/r
tanθ = y/x
so, if secθ = -2 in QIII (where both x and y are negative)
x = -1
r = 2
y = -√3
cotθ = x/y = -1/-√3 = 1/√3