# Is it possible to get some help? Thanks!

## Apr 28, 2018

$\tan \theta = \frac{7 \sqrt{15}}{64}$ $\sec \theta = \frac{- 8 \sqrt{15}}{15}$

#### Explanation:

We're told that $\sin \theta$ is negative; however, $\cot \theta = \cos \frac{\theta}{\sin} \theta$ is positive.

This tells us that both sine and cosine must be negative (the negatives cancel out to give a positive cotangent), so we're working in the fourth quadrant.

Then, $\tan \theta = \sin \frac{\theta}{\cos} \theta$ will also be positive.

Furthermore, since $\sec \theta = \frac{1}{\cos} \theta , \sec \theta$ will also be negative as a result of the negative cosine.

Keeping this in mind, we'll continue.

Recall the identity

${\sin}^{2} \theta + {\cos}^{2} \theta = 1$

$\sin \theta = \left(- \frac{7}{8}\right) , {\sin}^{2} \theta = {\left(- \frac{7}{8}\right)}^{2} = \frac{49}{64}$

Thus,

$\frac{49}{64} + {\cos}^{2} \theta = \frac{64}{64}$

${\cos}^{2} \theta = \frac{64 - 49}{64}$

${\cos}^{2} \theta = \frac{15}{64}$

$\cos \theta = \pm \sqrt{\frac{15}{64}} = \pm \frac{\sqrt{15}}{8}$

We want the negative result:

$\cos \theta = - \frac{\sqrt{15}}{8}$

Then,

$\sec \theta = \frac{1}{\cos} \theta = \frac{1}{\frac{\sqrt{15}}{8}} = - \frac{8}{\sqrt{15}}$

Rationalizing the denominator (multiply by $\frac{\sqrt{15}}{\sqrt{15}}$), we get

$\sec \theta = - \frac{8 \sqrt{15}}{15}$

We can now find the tangent:

$\tan \theta = \sin \frac{\theta}{\cos} \theta$

$\tan \theta = \frac{\cancel{-} \frac{7}{8}}{\cancel{-} \frac{8 \sqrt{15}}{15}}$

$\tan \theta = \frac{7}{8} \cdot \frac{15}{8 \sqrt{15}}$

$\tan \theta = \frac{105}{64 \sqrt{15}}$

Rationalizing, we get

$\tan \theta = \frac{105 \sqrt{15}}{960}$

$\tan \theta = \frac{7 \sqrt{15}}{64}$