Chapter 7.2, Problem 22AYU

To determine

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

Answer to Problem 22AYU

The exact value is $\frac{5\pi }{6}$ .

Explanation of Solution

Given information:

  ${\tan }^{-1}\left(\cot \frac{2\pi }{3}\right)$ .

Formula used:

Trigonometric Ratios :

  $\cot \left(\pi -\theta \right)=-\cot \theta$ .

Calculation:

Consider ,

  ${\tan }^{-1}\left(\cot \frac{2\pi }{3}\right)$

Let

  $\theta =\cot \frac{2\pi }{3}$

where $\frac{-\pi }{2}<\theta <\frac{\pi }{2}$ .

It follows $\frac{\pi }{2}<\theta \le \pi$ .

This implies $\theta$ lies in quadrant II.

Now,

  $\begin{array}{l}\cot \frac{2\pi }{3}=\cot \left(\pi -\frac{\pi }{3}\right)=-\cot \frac{\pi }{3}\\ =-\cot \frac{\pi }{3}=-\left(\frac{1}{\sqrt{3}}\right)\\ =-\frac{1}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=-\frac{\sqrt{3}}{3}\end{array}$

Where ${\tan }^{-1}\theta$

  $\frac{\pi }{2}\le \theta \le \pi$

and the negative value lies between zero and $\frac{\pi }{2}$ .

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

  $\begin{array}{l}=\tan \theta =-\frac{1}{\sqrt{3}}\\ =\tan \theta =-\frac{\pi }{6}\end{array}$

  $=\left(\pi -\frac{\pi }{6}\right)=\frac{5\pi }{6}$

Hence , the exact value of ${\tan }^{-1}\left(\cot \frac{2\pi }{3}\right)$ is $\frac{5\pi }{6}$ .