angles α and β are in standard position and intersect the unit circle at points a and e, respectively. the…

Answer:

A. $g(\alpha)<g(\beta)$ and $g(\alpha)<0$

Explanation:

Step1: Analyze the tangent - function's properties

The tangent function $y = \tan x$ has a period of $\pi$ and is increasing on each interval $\left(-\frac{\pi}{2}+k\pi,\frac{\pi}{2}+k\pi\right),k\in\mathbb{Z}$. In the interval $\left(\frac{3\pi}{2},2\pi\right)$, the tangent function $y = \tan x$ is increasing.

Step2: Compare $g(\alpha)$ and $g(\beta)$

Since $\frac{3\pi}{2}<\alpha<\beta < 2\pi$ and $y = \tan x$ is increasing on $\left(\frac{3\pi}{2},2\pi\right)$, we have $g(\alpha)=\tan\alpha<g(\beta)=\tan\beta$.

Step3: Determine the sign of $g(\alpha)$

The tangent function $y = \tan x=\frac{\sin x}{\cos x}$. In the fourth - quadrant ($\frac{3\pi}{2}<x < 2\pi$), $\sin x<0$ and $\cos x>0$. So, for $\alpha\in\left(\frac{3\pi}{2},2\pi\right)$, $g(\alpha)=\tan\alpha=\frac{\sin\alpha}{\cos\alpha}<0$.