Why is "secant" in the third quadrant negative?

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Why is "secant" in the third quadrant negative?

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Answer 1 · 2018-04-27T06:46:20

By definintion

Explanation:

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

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

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Why is "secant" in the third quadrant negative?

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