Chapter 4.3, Problem 58E

(a)

To determine

The value of $\cot \beta$ with the help of the trigonometric identities.

Answer to Problem 58E

The value of $\cot \beta$ is $\frac{1}{3}$ .

Explanation of Solution

Given Information:

The value of $\tan \beta =3$ .

Formula Used:

Use the trigonometric identity $\cot \beta =\frac{1}{\tan \beta }$

Calculation:

Substitute $3$ for $\tan \beta$ in above identity.

  $\begin{array}{c}\cot \beta =\frac{1}{\tan \beta }\\ =\frac{1}{3}\end{array}$

Therefore the value of $\cot \beta$ is $\frac{1}{3}$ .

(b)

To determine

The value of $\cos \beta$ with the help of the trigonometric identities.

Answer to Problem 58E

The value of $\cos \beta$ is $\frac{\sqrt{10}}{10}$ .

Explanation of Solution

Given Information:

The value of $\tan \beta =3$ .

Formula Used:

Trigonometric identity ${\sec }^2\beta =1+{\tan }^2\beta$

  $\cos \beta =\frac{1}{\sec \beta }$

Calculation:

Substitute $3$ for $\tan \beta$ in above identity.

  $\begin{array}{c}{\sec }^2\beta =1+{\tan }^2\beta \\ =1+{\left(3\right)}^2\\ =1+9\\ =10\end{array}$ ,

Substitute $\sqrt{10}$ for the value of $\sec \beta$ in the identity.

  $\begin{array}{c}\cos \beta =\frac{1}{\sec \beta }\\ =\frac{1}{\sqrt{10}}\\ =\frac{\sqrt{10}}{10}\end{array}$

Therefore the value of $\cos \beta$ is $\frac{\sqrt{10}}{10}$ .

(c)

To determine

The value of $\tan \left(90°-\beta \right)$ with the help of the trigonometric identities.

Answer to Problem 58E

The value of $\tan \left(90°-\beta \right)$ is $\frac{1}{3}$ .

Explanation of Solution

Given Information:

The value of $\tan \beta =3$ .

Formula Used:

Trigonometric identity $\tan \left(90°-\beta \right)=\cot \beta$

  $\cot \beta =\frac{1}{\tan \beta }$

Calculation:

Substitute $3$ for $\tan \beta$ in above identity.

  $\begin{array}{c}\cot \beta =\frac{1}{\tan \beta }\\ =\frac{1}{3}\end{array}$

Substitute $\frac{1}{3}$ for $\cot \beta$ in the identity.

  $\begin{array}{c}\tan \left(90°-\beta \right)=\cot \beta \\ =\frac{1}{3}\end{array}$

Therefore the value of $\tan \left(90°-\beta \right)$ is $\frac{1}{3}$ .

(d)

To determine

The value of $\csc \beta$ with the help of the trigonometric identities.

Answer to Problem 58E

The value of $\csc \beta$ is $\frac{\sqrt{10}}{3}$ .

Explanation of Solution

Given Information:

The value of $\tan \beta =3$ .

Formula Used:

Trigonometric identity $1+{\cot }^2\beta ={\csc }^2\beta$

  $\cot \beta =\frac{1}{\tan \beta }$

Calculation:

Substitute $3$ for $\tan \beta$ in above identity.

  $\begin{array}{c}\cot \beta =\frac{1}{\tan \beta }\\ =\frac{1}{3}\end{array}$

Substitute $\frac{1}{3}$ for the value of $\cot \beta$ in the identity.

  $\begin{array}{c}{\csc }^2\beta =1+{\cot }^2\beta \\ =1+{\left(\frac{1}{3}\right)}^2\\ =1+\frac{1}{9}\\ =\frac{10}{9}\end{array}$

Therefore the value of $\csc \beta$ is $\frac{\sqrt{10}}{3}$ .