Question

Why is "secant" in the third quadrant negative?

Answer 1

By definintion

Explanation:

If we think of a right triangle with vertical leg `y` and horizontal leg `x` and hypotenuse `r` with an angle `theta` where `x` and `r` intersect, we know that:
`x=rcos(theta)`
and
`y=rsin(theta)`.

`sec(x) = 1/cos(x)` by definition, that is `sec(x) = r/x`
Since `r` is a radius, it must be positive, so `sec(x)` is negative anywhere `x` is negative. This is in Quadrants `II` and `III`.