Chapter 7.2, Problem 11AYU

To determine

To calculate: The exact value of expression $\tan \left[{\cos }^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right]$ .

Answer to Problem 11AYU

The exact value is $-\frac{1}{\sqrt{3}}$ .

Explanation of Solution

Given information:

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

Formula used:

Equation $y={\cos }^{-1}x$ means $\cos y=x$ .

Theorem :- For any angle $\theta$ , in standard position, let $P=(x,y)$ be the point on the terminal side of $\theta$ , that is on the circle $x^2+y^2=r^2$ . Then

  $\begin{array}{l}\sin \theta =\frac{y}{r}\\ \cos \theta =\frac{x}{r}\\ \tan \theta =\frac{y}{x},x\ne 0\end{array}$$\begin{array}{l}\cos ec\theta =\frac{r}{y},y\ne 0\\ \sec \theta =\frac{r}{x},x\ne 0\\ \cot \theta =\frac{x}{y},y\ne 0\end{array}$ .

Calculation:

Consider ,

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

Let

  $\begin{array}{l}\theta ={\cos }^{-1}\left(-\frac{\sqrt{3}}{2}\right)\\ \cos \theta =-\frac{\sqrt{3}}{2}\end{array}$

where $-\pi <\theta <\pi$ .

Since $\cos \theta <0$ , it follows $\frac{\pi }{2}<\theta \le \pi$ .

This implies $\theta$ lies in quadrant II.

Now,

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

  $=x=-\sqrt{3};r=2$

Using equation of circle we have,

  $\begin{array}{l}=x^2+y^2=r^2\\ ={(}^{-}+y^2={\left(2\right)}^2\\ =3+y^2=4\\ =y^2=4-3\\ =y^2=1\\ =y=\pm 1\end{array}$

Thus we have,

  $=x=-\sqrt{3};y=1;r=2$ .

Now,

  $\tan \left[{\cos }^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right]=\tan \theta =\frac{-1}{\sqrt{3}}$ ( where $\tan \frac{y}{x}$ )

Hence , the exact value of $\tan \left[{\cos }^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right]$ is $-\frac{1}{\sqrt{3}}$ .