# If cos theta = -8/9, and theta is in Quadrant III, how do you find tanthetacottheta+csctheta?

May 10, 2017

See below. Answer: $\frac{17 - 9 \sqrt{17}}{17}$

#### Explanation:

Original question: Given $\cos \theta = - \frac{8}{9}$, $\theta$ in Quadrant III, find $\tan \theta \cot \theta + \csc \theta$

First, we notice that the question is really only asking us to find $\sin \theta$ since $\tan \theta \cot \theta + \csc \theta = \left(\sin \frac{\theta}{\cos} \theta\right) \left(\cos \frac{\theta}{\sin} \theta\right) + \frac{1}{\sin} \theta = 1 + \frac{1}{\sin} \theta$, making this problem a lot less complicated than it seems.

We also know that since $\theta$ is in Quadrant III, both $\cos \theta$ and $\sin \theta$ are negative.

Consider Pythagorean's identity, ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$.

Since we know that $\cos \left(\theta\right) = - \frac{8}{9}$, we can plug in this value into Pythagorean's identity to find $\sin \left(\theta\right)$:
${\sin}^{2} \theta + {\left(- \frac{8}{9}\right)}^{2} = 1$
${\sin}^{2} \theta + \frac{64}{81} = 1$
${\sin}^{2} \theta = \frac{17}{81}$
$\sin \theta = \pm \sqrt{\frac{17}{81}} = \pm \frac{\sqrt{17}}{9}$
However, since we know that $\sin \theta$ is in Quadrant III and must be negative, then $\sin \theta = - \frac{\sqrt{17}}{9}$

Finally, since we are solving for $1 + \csc \theta$ or $1 + \frac{1}{\sin} \theta$, our final answer is $1 + \frac{1}{- \frac{\sqrt{17}}{9}} = 1 - \frac{9}{\sqrt{17}} = \frac{17 - 9 \sqrt{17}}{17}$.

Answer: $\frac{17 - 9 \sqrt{17}}{17}$