Question

How do you use special angles to find the ratios of sec 180 degrees?

Answer 1

`sec(180^@) = -1`

Explanation:

`180^@` has a reference angle of `0^@` and falls along the negative X-axis.

#sec= 1/cos = ("hypotenuse")/("adjacent")

`color(white)("XXX")`the `"hypotenuse"` is always positive.
`color(white)("XXX")`the `"adjacent"` is the X-axis value and along the negative X-axis it is negative.

For an angle of `0^@` the lengths of the `"hypotenuse"` and `"adjacent"` sides are equal.

Therefore `sec(180^@) = -sec(0^@) = -1`