Chapter 7.2, Problem 27AYU

To determine

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

Answer to Problem 27AYU

The exact value is $\frac{\sqrt{5}}{2}$ .

Explanation of Solution

Given information:

  $\sec \left({\tan }^{-1}\frac{1}{2}\right)$ .

Formula used:

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

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 ,

  $\sec \left({\tan }^{-1}\frac{1}{2}\right)$

Let

  $\begin{array}{l}\theta ={\tan }^{-1}\frac{1}{2}\\ \tan \theta =\frac{1}{2}\end{array}$

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

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

This implies $\theta$ lies in quadrant I.

Now,

  $\tan \theta =\frac{y}{x}=\frac{1}{2}$

  $=y=1;x=2$

Using equation of circle we have,

  $\begin{array}{l}=x^2+y^2=r^2\\ ={\left(2\right)}^2+{\left(1\right)}^2=r^2\\ =4+1=r^2\\ =5=r^2\\ =r=\pm \sqrt{5}\end{array}$

Thus we have,

  $=x=2;y=1;r=\sqrt{5}$ .

Now,

  $\sec \left({\tan }^{-1}\frac{1}{2}\right)=\sec \theta =\frac{\sqrt{5}}{2}$ ( where $\sec \frac{r}{x}$ )

Hence , the exact value of $\sec \left({\tan }^{-1}\frac{1}{2}\right)$ is $\frac{\sqrt{5}}{2}$ .