Chapter 4, Problem 88RE

To determine

To Find:Exact value oftrigonometric expression $\sec (\arctan \frac{12}{5})$ .

Answer to Problem 88RE

The exact value of trigonometric expression $\sec (\arctan \frac{12}{5})$ is $\frac{13}{5}=2.6$ .

Explanation of Solution

Given information:

To draw right angle triangle.

Formula used:

Pythagoras theorem that is $p^2+b^2=h^2$ to find height and trigonometric formula $\begin{array}{l}\sec ({\sec }^{-1}x)=x\\ and{\cos }^{-1}x={\sec }^{-1}\frac{1}{x}\end{array}$

Where p is perpendicular is base,h is height.

Calculation:

Consider the statement “given trigonometric expression”

Recall that by the Pythagoras formula on right angle triangle

  

Therefore, given that perpendicular is $12$

Base is $5$

Height is h

So by Pythagoras formula,

  $\begin{array}{l}\sqrt{p^2+b^2}=h\\ \sqrt{{12}^2+5^2}=h\\ \sqrt{144+25}=h\\ \sqrt{169}=h\\ 13=h\end{array}$

Where h is height

Now,recall the trigonometric formula of ${\cos }^{-1}x={\sec }^{-1}\frac{1}{x}$

Since, perpendicular is $12$

Base is $5$

Height is $13$

So, $\cos x=\frac{5}{13}$

  $x={\cos }^{-1}\frac{5}{13}$

Therefore, ${\cos }^{-1}\frac{5}{13}={\sec }^{-1}\frac{13}{5}$

Recall the trigonometric formula of $\sec ({\sec }^{-1}x)=x$

Therefore, from right angle $\sec ({\sec }^{-1}\frac{13}{5})=\frac{13}{5}=2.6$

So, the exact value of trigonometric function $\sec ({\sec }^{-1}\frac{13}{5})is2.6$