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

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