# How do you solve csc[tan^-1 (-2)] ?

Dec 3, 2016

$\csc \left({\tan}^{- 1} \left(- 2\right)\right) = - \frac{\sqrt{5}}{2}$

#### Explanation:

From the definition of inverse ratios, if ${\tan}^{- 1} \left(- 2\right) = \theta$, then $\tan \theta = - 2$

As $\tan \theta = - 2$, we have $\cot \theta = - \frac{1}{2}$ and

$\csc \theta = - \sqrt{1 + {\left(- \frac{1}{2}\right)}^{2}} = - \sqrt{1 + \frac{1}{4}}$

= $- \sqrt{\frac{5}{4}} = - \frac{\sqrt{5}}{2}$

Note that as $\tan \theta$ is negative so would be $\csc \theta$ as domain for $\theta$ is $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$.

hence $\csc \theta = \csc \left({\tan}^{- 1} \left(- 2\right)\right) = - \frac{\sqrt{5}}{2}$