If `sec x > 0` and `cot x < 0` in which quadrant does the terminal side of position angle x lie?

The endpoint of an angle `x` lies in the 4th quadrant. Here is why.

First of all, let's recall a few definitions.

Recall the definition of a trigonometric function `sec(x)`:
`sec(x) =` (by definition) `=1/cos(x)`

Recall the definition of a trigonometric function `cot(x)`:
`cot(x) =` (by definition) `=cos(x)/sin(x)`

Recall the definition of a trigonometric function `cos(x)`:
Consider a unit circle around the origin of coordinate `O` and a point `A` on this circle such that an angle from the positive direction of the X-axis `OX` counterclockwise to a ray `OA` equals to `x` (usually, radians). Then the abscissa (X-coordinate) of the point `A` is, by definition, `cos(x)`.

Finally, recall the definition of a trigonometric function `sin(x)`:
The same arrangement as above for `cos(x)`, but in this case the ordinate (Y-coordinate) of the point `A` is, by definition, `sin(x)`.

Since `sec(x)=1/cos(x)>0`, `cos(x)>0`, which means that abscissa of a position of the endpoint of an angle `x` is positive.

Since `cot(x)=cos(x)/sin(x) < 0` and `cos(x)>0`, `sin(x)` must be negative, which means that ordinate of a position of the endpoint of an angle `x` is negative.

So, the endpoint of an angle `x` has positive abscissa (X-coordinate) and negative ordinate (Y-coordinate), which puts it into 4th quadrant.