# Given Tangent Theta= -4/5, Theta in quadrant 2 , how do you find Cos Theta?

Jul 12, 2016

$\cos \theta = - \frac{5}{\sqrt{41}}$.

#### Explanation:

Given that $\tan \theta = - \frac{4}{5} , \theta \in {Q}_{I I}$

Recall that $\cos \theta = \frac{1}{\sec} \theta$ and, $\tan$ and $\sec$ are related

by the Identity $: {\sec}^{2} \theta = 1 + {\tan}^{2} \theta .$

$\therefore {\sec}^{2} \theta = 1 + {\left(- \frac{4}{5}\right)}^{2} = 1 + \frac{16}{25} = \frac{41}{25.}$

$\therefore \sec \theta = \pm \frac{\sqrt{41}}{5} \Rightarrow \cos \theta = \pm \frac{5}{\sqrt{41}}$

But, $\theta \in {Q}_{I I} \Rightarrow \cos \theta$ is $- v e$.

Hence, $\cos \theta = - \frac{5}{\sqrt{41}}$.