# Question #47acb

Mar 16, 2017

Let's write $\sec \left(\theta\right) = - 4.5$ as $\sec \left(\theta\right) = - \frac{9}{2}$

Then:

$\cos \left(\theta\right) = \frac{1}{\sec} \left(\theta\right)$

$\cos \left(\theta\right) = - \frac{2}{9}$

And:

${\sin}^{2} \left(\theta\right) = 1 - {\left(- \frac{2}{9}\right)}^{2} \text{ [1]}$

When we take square root of both sides of equation [1], we must use $\pm$ on the right:

$\sin \left(\theta\right) = \pm \sqrt{1 - {\left(\frac{2}{9}\right)}^{2}}$

We are given that $\theta$ is the second quadrant and we know that the sine function is positive in the second quadrant so we drop the $\pm$, thereby, indicating only the positive value:

$\sin \left(\theta\right) = \sqrt{1 - {\left(\frac{2}{9}\right)}^{2}}$

$\sin \left(\theta\right) = \sqrt{\frac{81}{81} - \frac{4}{81}}$

$\sin \left(\theta\right) = \sqrt{\frac{77}{81}}$

$\sin \left(\theta\right) = \frac{\sqrt{77}}{9}$

$\tan \left(\theta\right) = \sin \frac{\theta}{\cos} \left(\theta\right)$

$\tan \left(\theta\right) = \frac{\frac{\sqrt{77}}{9}}{- \frac{2}{9}}$

$\tan \left(\theta\right) = - \frac{\sqrt{77}}{2}$

$\cot \left(\theta\right) = \frac{1}{\tan} \left(\theta\right)$

$\cot \left(\theta\right) = - \frac{2}{\sqrt{77}} = - 2 \frac{\sqrt{77}}{77}$

$\csc \left(\theta\right) = \frac{1}{\sin} \left(\theta\right)$

$\csc \left(\theta\right) = \frac{9}{\sqrt{77}}$

$\csc \left(\theta\right) = 9 \frac{\sqrt{77}}{77}$