Question

In Problems 9–36, find the exact value of each expression.

$\tan \left[{\cos }^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right]$

Short answer

The value of the expression$\tan \left[{\cos }^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right]=-\frac{\sqrt{3}}{3}$

Step-by-step solution

Step 1. Given information.

The given expression is:

$\tan \left[{\cos }^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right]$

Let's take $\theta ={\cos }^{-1}\left(-\frac{\sqrt{3}}{2}\right)$,

$\cos \theta =-\frac{\sqrt{3}}{2}$

$0\le \theta \le \mathrm{\pi }$is the domain of the function.

Step 2. Find the value of 

Initially, we have${\cos }^{-1}$ and the bounds of cosfunction are going to be in the interval $\left[0,\mathrm{\pi }\right]$. It represents the range of ${\cos }^{-1}$.

By using the unit circle, we can say that $\theta$must be equal to $\frac{5\mathrm{\pi }}{6}$, so that we can get $-\frac{\sqrt{3}}{2}$.

$$\begin{gathered} \cos \theta =-\frac{\sqrt{3}}{2} \\ {\cos }^{-1}\left(-\frac{\sqrt{3}}{2}\right)=\frac{5\mathrm{\pi }}{6} \end{gathered}$$

Step 3. Find the exact value of the expression.

$$\begin{gathered} \tan \left(\frac{5\mathrm{\pi }}{6}\right)=-\frac{1}{\sqrt{3}} \\ =-\frac{\sqrt{3}}{3} \end{gathered}$$

Therefore,$\tan \left[{\cos }^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right]=-\frac{\sqrt{3}}{3}$.