Question

question 20 (5 points) listen find the value of tan ( sin ^ { - 1 } ( - \frac { 1 } { 2 } ) ) . \sqrt { 3 } \frac { \sqrt { 3 } } { 3 } - \frac { \sqrt { 3 } } { 3 } - \sqrt { 3 }

Explanation:

Step1: Let $\theta=\sin^{-1}(-\frac{1}{2})$

By the definition of the inverse - sine function, $\sin\theta =-\frac{1}{2}$ and $\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]$. So, $\theta =-\frac{\pi}{6}$.

Step2: Calculate $\tan\theta$

Substitute $\theta =-\frac{\pi}{6}$ into $\tan\theta$. We know that $\tan\theta=\tan(-\frac{\pi}{6})$. Using the property $\tan(-x)=-\tan x$, we have $\tan(-\frac{\pi}{6})=-\tan\frac{\pi}{6}$. And since $\tan\frac{\pi}{6}=\frac{\sqrt{3}}{3}$, then $\tan(-\frac{\pi}{6})=-\frac{\sqrt{3}}{3}$.

Answer:

$-\frac{\sqrt{3}}{3}$ (corresponding to the third option)