find the exact value of $\tan^{-1}left(-\frac{sqrt{3}}{3}\right)$. $\tan^{-1}left(-\frac{sqrt{3}}{3}\right)=s…
find the exact value of $\tan^{-1}left(-\frac{sqrt{3}}{3}\right)$. $\tan^{-1}left(-\frac{sqrt{3}}{3}\right)=square$ (type your answer in radians.)
Answer
Explanation:
Step1: Recall the range of inverse - tangent function
The range of $y = \tan^{-1}(x)$ is $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$.
Step2: Find the angle whose tangent is $-\frac{\sqrt{3}}{3}$
We know that $\tan\left(-\frac{\pi}{6}\right)=-\frac{\sqrt{3}}{3}$ and $-\frac{\pi}{6}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$.
Answer:
$-\frac{\pi}{6}$