Chapter 7, Problem 27RE

To determine

To Calculate: The simplified value of the expression $\sec \left({\tan }^{-1}\left(\frac{\sqrt{3}}{3}\right)\right)$ .

Answer to Problem 27RE

The simplified value of the expression $\sec \left({\tan }^{-1}\left(\frac{\sqrt{3}}{3}\right)\right)$ is $\frac{2}{\sqrt{3}}$.

Explanation of Solution

Given information:

The expression is $\sec \left({\tan }^{-1}\left(\frac{\sqrt{3}}{3}\right)\right)$ .

Formula used:

For the tangent function and its inverse, the following holds:

  $\begin{array}{c}f\left(f^{-1}\left(x\right)\right)=\tan \left({\tan }^{-1}\left(x\right)\right)\\ =x\end{array}$

Where, $-\infty \le x\le \infty$

Tangent function and secant function related as,

  $\sec \theta =\sqrt{\left({\tan }^2\theta +1\right)}$

Calculation:

Consider the expression $\sec \left({\tan }^{-1}\left(\frac{\sqrt{3}}{3}\right)\right)$ .

Recall that tangent function and secant function related as,

  $\sec \theta =\sqrt{\left({\tan }^2\theta +1\right)}$

Using it,

  $\sec \left({\tan }^{-1}\left(\frac{\sqrt{3}}{3}\right)\right)=\sqrt{\left({\tan }^2\left({\tan }^{-1}\left(\frac{\sqrt{3}}{3}\right)\right)+1\right)}$

Observe that the value $\frac{\sqrt{3}}{3}$ is lies between $\infty$ and $\infty$ .

Recall that for the tangent function and its inverse, the following holds:

  $\begin{array}{c}f\left(f^{-1}\left(x\right)\right)=\tan \left({\tan }^{-1}\left(x\right)\right)\\ =x\end{array}$

Where, $-\infty \le x\le \infty$

Simplify further,

  $\begin{array}{c}\sqrt{\left({\tan }^2\left({\tan }^{-1}\left(\frac{\sqrt{3}}{3}\right)\right)+1\right)}=\sqrt{\left({\left(\tan \left({\tan }^{-1}\left(\frac{\sqrt{3}}{3}\right)\right)\right)}^2+1\right)}\\ =\sqrt{\left({\left(\frac{\sqrt{3}}{3}\right)}^2+1\right)}\\ =\sqrt{\left({\left(\frac{1}{\sqrt{3}}\right)}^2+1\right)}\\ =\frac{2}{\sqrt{3}}\end{array}$

Thus, the simplified value of the expression $\sec \left({\tan }^{-1}\left(\frac{\sqrt{3}}{3}\right)\right)$ is $\frac{2}{\sqrt{3}}$ .