Question

Θ is in Quadrant III and cos^2 0= 1/4

Answer 1

I think you mean to say $\cos^2\theta=\dfrac14$. If $\theta$ is in the third quadrant, then $\sin\theta$ and $\cos\theta$. Recall that


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

We expect $\sin\theta$ to be negative, so we take the negative root. We end up with

$\sin\theta=-\sqrt{1-\dfrac14}=-\dfrac{\sqrt3}2$

We also know to expect $\cos\theta$, so

$\cos\theta=\pm\sqrt{\dfrac14}\implies\cos\theta=-\dfrac12$


By definition, we have

$\cot\theta=\dfrac{\cos\theta}{\sin\theta}$

and so

$\cot\theta=\dfrac{-\frac12}{-\frac{\sqrt3}2}=\dfrac1{\sqrt3}$