# How do you determine tantheta given costheta=1/8, (3pi)/2<theta<2pi?

Dec 19, 2017

$- 3 \sqrt{7.}$

#### Explanation:

$\cos \theta = \frac{1}{8} \Rightarrow \sec \theta = 8.$

$\therefore {\tan}^{2} \theta = {\sec}^{2} \theta - 1 = {8}^{2} - 1 = 63 = 9 \cdot 7.$

Hence, $\tan \theta = \sqrt{63} = \pm 3 \sqrt{7.}$

But, given that, $\frac{3}{2} \pi < \theta < 2 \pi$,

$\Rightarrow \tan \theta = - 3 \sqrt{7.}$

Dec 19, 2017

$\tan \theta = - 3 \sqrt{7}$

#### Explanation:

$\cos \theta = \frac{1}{8}$

$\mathmr{and} , {\cos}^{2} \theta = \frac{1}{64}$

$\mathmr{and} , 1 - {\sin}^{2} \theta = \frac{1}{64}$$\textcolor{b l u e}{\left[\left[{\sin}^{2} + {\cos}^{2} \theta = 1 , s o , 1 - {\sin}^{2} \theta = {\cos}^{2} \theta\right]\right]}$

$\mathmr{and} , {\sin}^{2} \theta = 1 - \frac{1}{64}$

$\mathmr{and} , {\sin}^{2} \theta = \frac{63}{64}$

$\mathmr{and} , \sin \theta = \pm \frac{\sqrt{63}}{8}$

Here $\sin \theta = - \frac{\sqrt{63}}{8}$ ,As in fourth quarter the value of $\sin \theta$ is negative

$\tan \theta = \sin \frac{\theta}{\cos} \theta = \left(- \frac{\sqrt{63}}{8} / \frac{1}{8} = - \frac{\sqrt{63}}{\cancel{8}} \times \cancel{8}\right) = - \sqrt{63}$

So,$\tan \theta = - \sqrt{63} = - 3 \sqrt{7}$