# How do you use csctheta=5 to find tantheta?

May 2, 2017

$\tan \left(\theta\right) = \pm \frac{\sqrt{6}}{12}$

#### Explanation:

There are several methods to solve this problem.

Geometric
Imagine a right-angle triangle with an angle $\theta$. Since $\csc \left(\theta\right) = 5$, the hypotenuse is $5$, its opposite side is $1$, and its adjacent side is $\sqrt{{5}^{2} - {1}^{2}} = \sqrt{24} = 2 \sqrt{6}$. Thus, $\tan \left(\theta\right)$ is the opposite over the adjacent, or $\tan \left(\theta\right) = \frac{1}{2 \sqrt{6}} = \frac{\sqrt{6}}{12}$.

Now, since $\csc \left(\theta\right) = 5 > 0$, $\theta$ is either in quadrant 1 or 2. Because $\tan \left(\theta\right)$ is positive in quadrant 1 and negative in quadrant 2, there are two possible answers: $\pm \frac{\sqrt{6}}{12}$.

Another way to realize that there is a second solution is to realize that it is possible to draw the right triangle in the second quadrant. The hypotenuse is $5$ (remember that the hypotenuse is always positive). The opposite side, which is the $y$-value, is still positive (it is still $1$). Thus, $\csc \left(\theta\right) = \frac{5}{1} = 5$. However, the adjacent side, which is the $x$-value, becomes negative in quadrant 2 ($- \sqrt{{5}^{2} - {1}^{2}} = - \sqrt{24} = - 2 \sqrt{6}$). Thus, $\tan \left(\theta\right)$, being the opposite over the adjacent, has the exact same value but with a negative sign.

Alegbraic
Another method is algebraic:
$\csc \left(\theta\right) = 5$
$\frac{1}{\sin} \left(\theta\right) = 5$
$\sin \left(\theta\right) = \frac{1}{5}$
$\theta = \arcsin \left(\frac{1}{5}\right)$
$\tan \left(\theta\right) = \tan \left(\arcsin \left(\frac{1}{5}\right)\right) = \sin \frac{\arcsin \left(\frac{1}{5}\right)}{\cos} \left(\arcsin \left(\frac{1}{5}\right)\right)$
$\tan \left(\theta\right) = \frac{\frac{1}{5}}{\pm \sqrt{1 - {\sin}^{2} \left(\arcsin \left(\frac{1}{5}\right)\right)}} = \frac{1}{\pm 5 \sqrt{1 - {\left(\frac{1}{5}\right)}^{2}}}$
$\tan \left(\theta\right) = \frac{1}{\pm \sqrt{24}} = \pm \frac{\sqrt{6}}{12}$