Chapter 4.3, Problem 57E

(a)

To determine

The value of $\tan \alpha$ with the help of the trigonometric identities.

Answer to Problem 57E

The value of $\tan \alpha$ is $\frac{1}{4}$ .

Explanation of Solution

Given Information:

The value of $\cot \alpha =4$ .

Formula Used:

Trigonometric identity $\tan \alpha =\frac{1}{\cot \alpha }$ .

Calculation:

Substitute $4$ for $\cot \alpha$ in above identity.

  $\begin{array}{c}\tan \alpha =\frac{1}{\cot \alpha }\\ =\frac{1}{4}\end{array}$

Therefore the value of $\tan \alpha$ is $\frac{1}{4}$ .

(b)

To determine

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

Answer to Problem 57E

The value of $\csc \alpha$ is $\sqrt{17}$ .

Explanation of Solution

Given Information:

The value of $\cot \alpha =4$ .

Formula Used:

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

Calculation:

Substitute $4$ for $\cot \alpha$ in above identity.

  $\begin{array}{c}{\csc }^2\alpha =1+{\cot }^2\alpha \\ =1+{\left(4\right)}^2\\ =1+16\\ =17\end{array}$

Therefore the value of $\csc \alpha$ is $\sqrt{17}$ .

(c)

To determine

The value of $\sec \alpha$ with the help of the trigonometric identities.

Answer to Problem 57E

The value of $\sec \alpha$ is $\frac{\sqrt{17}}{4}$ .

Explanation of Solution

Given Information:

The value of $\cot \alpha =4$ .

Formula Used:

Trigonometric identity $1+{\tan }^2\alpha ={\sec }^2\alpha$ , where $\tan \alpha =\frac{1}{\cot \alpha }$

Calculation:

Substitute $4$ for $\cot \alpha$ in identity $\tan \alpha =\frac{1}{\cot \alpha }$ .

  $\tan \alpha =\frac{1}{4}$

Now substitute $\frac{1}{4}$ for $\tan \alpha$ in the above identity.

  $\begin{array}{c}{\sec }^2\alpha =1+{\tan }^2\alpha \\ =1+{\left(\frac{1}{4}\right)}^2\\ =1+\frac{1}{16}\\ =\frac{17}{16}\end{array}$

Therefore the value of $\sec \alpha$ is $\frac{\sqrt{17}}{4}$ .

(d)

To determine

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

Answer to Problem 57E

The value of $\tan \left(90°-\alpha \right)$ is $4$ .

Explanation of Solution

Given Information:

The value of $\cot \alpha =4$ .

Formula Used:

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

Calculation:

Substitute $4$ for $\cot \alpha$ in above identity.

  $\begin{array}{c}\tan \left(90°-\alpha \right)=\cot \alpha \\ =4\end{array}$

Therefore the value of $\tan \left(90°-\alpha \right)$ is $4$ .