Question

4 multiple choice 1 point which of the following is $tan(\frac{4\pi}{6})$? $-\sqrt{3}$ $-\frac{\sqrt{3}}{2}$ $-\frac{\sqrt{3}}{3}$ $-\frac{1}{2}$ previous

Explanation:

Step1: Use the tangent subtraction formula

$$\tan(A - B)=\frac{\tan A-\tan B}{1 + \tan A\tan B}$$
We know that \(\frac{4\pi}{6}=\pi-\frac{\pi}{3}\), so \(\tan(\frac{4\pi}{6})=\tan(\pi-\frac{\pi}{3})\)
According to the formula \(\tan(\pi - \alpha)=-\tan\alpha\), here \(A = \pi\), \(B=\frac{\pi}{3}\), and \(\tan\pi = 0\)
\(\tan(\pi-\frac{\pi}{3})=\frac{\tan\pi-\tan\frac{\pi}{3}}{1+\tan\pi\tan\frac{\pi}{3}}=\frac{0 - \sqrt{3}}{1+0\times\sqrt{3}}=-\sqrt{3}\)

Step2: Another way: Unit - circle and reference - angle

The angle \(\theta=\frac{4\pi}{6}=\frac{2\pi}{3}\) (simplified). The reference angle \(\theta_{r}=\pi-\frac{2\pi}{3}=\frac{\pi}{3}\)
In the second quadrant (\(\frac{\pi}{2}<\frac{2\pi}{3}<\pi\)), the tangent function \(\tan\theta=\frac{\sin\theta}{\cos\theta}\) is negative.
We know that \(\tan\frac{\pi}{3}=\sqrt{3}\), so \(\tan(\frac{4\pi}{6})=-\sqrt{3}\)

Answer:

\(-\sqrt{3}\)