# How do you determine tantheta given cottheta=-sqrt5/2,pi/2<theta<pi?

Jan 24, 2017

$\tan \theta = - \frac{2}{\sqrt{5}}$; $\sin \theta = - \frac{2}{3}$; $\csc \theta = - \frac{3}{2}$;$\cos \theta = - \frac{\sqrt{5}}{3}$;$\sec \theta = - \frac{3}{\sqrt{5}}$

#### Explanation:

Since $\tan \theta = \frac{1}{\cot} \theta$, you get:

$\tan \theta = - \frac{2}{\sqrt{5}}$

Since $\sin \theta = \pm \tan \frac{\theta}{\sqrt{1 + {\tan}^{2} \theta}}$ and sine is positive in the second quadrant:

sintheta=+(-2/sqrt5)/sqrt(1+(-2/sqrt5)^2

$= \frac{- \frac{2}{\sqrt{5}}}{\sqrt{1 + \frac{4}{5}}} = \frac{- \frac{2}{\sqrt{5}}}{\sqrt{\frac{9}{5}}} = - \frac{2}{\cancel{\sqrt{5}}} \cdot \frac{\cancel{\sqrt{5}}}{3} = - \frac{2}{3}$

Then $\csc \theta = \frac{1}{\sin} \theta = - \frac{3}{2}$

$\cos \theta = \pm \sqrt{1 - {\sin}^{2} \theta}$

In the second quadrant cosine is negative, then

$\cos \theta = - \sqrt{1 - {\left(- \frac{2}{3}\right)}^{2}} = - \sqrt{1 - \frac{4}{9}} = - \sqrt{\frac{5}{9}} = - \frac{\sqrt{5}}{3}$

and $\sec \theta = \frac{1}{\cos} \theta = - \frac{3}{\sqrt{5}}$