# Question #c43f5

May 31, 2017

$\sec \left(\theta\right) = - \frac{41}{80}$; ${180}^{\circ} < \theta < {270}^{\circ}$

#### Explanation:

We have: $\tan \left(\theta\right) = \frac{9}{40}$; ${180}^{\circ} < \theta < {270}^{\circ}$

Let's apply the Pythagorean identity $1 + {\tan}^{2} \left(\theta\right) = {\sec}^{2} \left(\theta\right)$:

$R i g h t a r r o w 1 + {\left(\frac{9}{40}\right)}^{2} = {\sec}^{2} \left(\theta\right)$

$R i g h t a r r o w 1 + \frac{81}{1600} = {\sec}^{2} \left(\theta\right)$

$R i g h t a r r o w \frac{1681}{1600} = {\sec}^{2} \left(\theta\right)$

$\therefore \sec \left(\theta\right) = \pm \frac{41}{80}$

The range of values of $\theta$ is given as ${180}^{\circ} < \theta < {270}^{\circ}$, i.e. the third quadrant.

This is the quadrant where all values of $\sec \left(\theta\right)$ are negative.

$\therefore \sec \left(\theta\right) = - \frac{41}{80}$

Jun 2, 2017

$\sec \theta = - \frac{41}{40}$

#### Explanation:

Use the identity

${\tan}^{2} \theta + 1 = {\sec}^{2} \theta$

We know that $\tan \theta = \frac{9}{40}$, so:

${\sec}^{2} \theta = {\left(\frac{9}{40}\right)}^{2} + 1$

${\sec}^{2} \theta = \frac{81}{1600} + 1$

${\sec}^{2} \theta = \frac{1681}{1600}$

$\sec \theta = \pm \frac{41}{40}$

To figure out whether the answer is positive or negative, we need to look at the quadrant that the angle is in. The problem tells us that the angle is between 180 and 270 degrees, so it must be in the third quadrant. In the third quadrant, $\sec \theta$ is negative since $\cos \theta$ is negative. Therefore, our single answer is:

$- \frac{41}{40}$