Answer
$-2$
Work Step by Step
RECALL:
(1)The reference angle of a given angle is equal to the smallest acute angle that the terminal side makes with the x-axis.
(2) Based on the location of the terminal side of an angle $\theta$, the reference angle can be found using the formula:
(i) Quadrant I: $\theta$
(ii) Quadrant II: $180^0-\theta$
(iii) Quadrant III: $\theta-180^o$
(iv) Quadrant IV: $360^o-\theta$
The given angle is in Quadrant III.
Use the formula in (iii) above to find its reference angle:
$=240^0-180^o
\\=60^o$
Note that $240^o$ is in Quadrant III, where secant is negative.
Thus,
$\sec{240^o} = -\sec{60^o}$
RECALL:
$\sec{\theta} = \dfrac{1}{\cos{\theta}}$
$60^o$ is a special angle whose cosine value is $0.5$.
Thus,
$\sec{240^o}
\\= -\sec{60^o}
\\=-\dfrac{1}{\cos{60^o}}
\\=-\dfrac{1}{0.5}
\\=-2$
