# How do you determine tantheta given sectheta=-5/4, 90^circ<theta<180^circ?

Nov 28, 2016

#### Explanation:

Here is how you derive a relationship between $\tan \left(\theta\right)$ and $\sec \left(\theta\right)$:

Start with ${\cos}^{2} \left(\theta\right) + {\sin}^{2} \left(\theta\right) = 1$

Divide both sides by ${\cos}^{2} \left(\theta\right)$:

$1 + {\sin}^{2} \frac{\theta}{\cos} ^ 2 \left(\theta\right) = \frac{1}{\cos} ^ 2 \left(\theta\right)$

Substitute ${\tan}^{2} \left(\theta\right)$ for ${\sin}^{2} \frac{\theta}{\cos} ^ 2 \left(\theta\right)$

$1 + {\tan}^{2} \left(\theta\right) = \frac{1}{\cos} ^ 2 \left(\theta\right)$

Substitute ${\sec}^{2} \left(\theta\right)$ for $\frac{1}{\cos} ^ 2 \left(\theta\right)$:

$1 + {\tan}^{2} \left(\theta\right) = {\sec}^{2} \left(\theta\right)$

Subtract 1 from both sides:

${\tan}^{2} \left(\theta\right) = {\sec}^{2} \left(\theta\right) - 1$

square root both sides:

$\tan \left(\theta\right) = \pm \sqrt{{\sec}^{2} \left(\theta\right) - 1}$

You do not need to remember this, because it is so easily derived.

Now that we have derived the equation for $\tan \left(\theta\right)$, substitute ${\left(- \frac{5}{4}\right)}^{2}$ for ${\sec}^{2} \left(\theta\right)$:

$\tan \left(\theta\right) = \pm \sqrt{{\left(- \frac{5}{4}\right)}^{2} - 1}$

Because we are told that the angle is in the second quadrant, we know that the tangent function is negative so we drop the + sign:

$\tan \left(\theta\right) = - \sqrt{{\left(- \frac{5}{4}\right)}^{2} - 1}$

$\tan \left(\theta\right) = - \sqrt{\frac{25}{16} - \frac{16}{16}}$

$\tan \left(\theta\right) = - \sqrt{\frac{9}{16}}$

$\tan \left(\theta\right) = - \frac{3}{4}$