Chapter 5.1, Problem 11E

To determine

To find the values of all six trigonometric functions by using given conditions.

Answer to Problem 11E

  $\sin x=\frac{2}{\sqrt{13}},\cos x=\frac{3}{\sqrt{13}},\csc x=\frac{\sqrt{13}}{2},\sec x=\frac{\sqrt{13}}{3},\tan x=\frac{2}{3}\&\cot x=\frac{3}{2}$

Explanation of Solution

Given:

  $\tan x=\frac{2}{3},\cos x>0$

Concept used:

Reciprocal identities:

  $\cot u=\frac{1}{\tan u},\csc u=\frac{1}{\sin u}$

Pythagorean identity:

  $\begin{array}{l}{\sin }^{2}u+{\cos }^{2}u=1\\ {\sec }^{2}u=1+{\tan }^{2}u\end{array}$

Calculation:

Given, $\tan x=\frac{2}{3},\cos x>0$

By using reciprocal identity, we have

  $\cot x=\frac{1}{\tan x}=\frac{1}{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=\frac{3}{2}$

Now by using Pythagorean identity, we have

  $\begin{array}{l}{\sec }^{2}x=1+{\tan }^{2}x\\ \Rightarrow {\sec }^{2}x=1+{\left(\frac{2}{3}\right)}^2\\ \Rightarrow {\sec }^{2}x=1+\frac{4}{9}\\ \Rightarrow {\sec }^{2}x=\frac{13}{9}\\ \Rightarrow \sec x=\pm \frac{\sqrt{13}}{3}\end{array}$

Since $\tan x>0\&\cos x>0$ , it follows that $x$ lies in quadrant I. Moreover, $\sec x$ is positive in quadrant I. Therefore,

  $\sec x=\frac{\sqrt{13}}{3}$

Now by using reciprocal identity, we have

  $\cos x=\frac{1}{\sec x}=\frac{1}{\raisebox{1ex}{$\sqrt{13}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=\frac{3}{\sqrt{13}}$

Now by using Pythagorean identity, we have

  $\begin{array}{l}{\sin }^{2}x=1-{\cos }^{2}x\\ \Rightarrow {\sin }^{2}x=1-{\left(\frac{3}{\sqrt{13}}\right)}^2\\ \Rightarrow {\sin }^{2}x=1-\frac{9}{13}\\ \Rightarrow {\sin }^{2}x=\frac{4}{13}\\ \Rightarrow \sin x=\pm \frac{2}{\sqrt{13}}\end{array}$

Since $\tan x>0\&\cos x>0$ , it follows that $x$ lies in quadrant I. Moreover, $\sin x$ is positive in quadrant I. Therefore,

  $\sin x=\frac{2}{\sqrt{13}}$

Now by using reciprocal identity, we have

  $\csc x=\frac{1}{\sin x}=\frac{1}{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$\sqrt{13}$}\right.}=\frac{\sqrt{13}}{2}$