Question

The angle θ lies in Quadrant III . cosθ=−1/6 What is tanθ ?

Answer 1

With θ in quadrant III, we expect to have both cosθ and sinθ negative. Then from the Pythagorean identity, we get

$\cos^2\theta+\sin^2\theta=1\implies\sin\theta=-\sqrt{1-\cos^2\theta}=-\dfrac{\sqrt{35}}6$

Then from the definition of tangent, that is

tanθ = sinθ / cosθ

we get

$\tan\theta=\dfrac{-\frac{\sqrt{35}}6}{-\frac16}=\boxed{\sqrt{35}}$

Answer 2

The value of $\tan \theta$ is required.

The required value in quadrant three is $\tan\theta=\sqrt{35}$

$\cos \theta=-\dfrac{1}{6}$

We have the identity

$\tan \theta=\dfrac{\sqrt{1-\cos^2\theta}}{\cos \theta}\\\Rightarrow \tan\theta=\dfrac{\sqrt{1-\left(-\dfrac{1}{6}\right)^2}}{-\dfrac{1}{6}}\\\Rightarrow \tan \theta=\dfrac{\sqrt{\dfrac{35}{36}}}{-\dfrac{1}{6}}\\\Rightarrow \tan\theta=\pm\sqrt{35}\times -1\\\Rightarrow \tan\theta=\pm\sqrt{35}$

The required value in quadrant three is $\tan\theta=\sqrt{35}$

Learn more:

https://brainly.com/question/11694897