Chapter 4.3, Problem 42E

(a)

To determine

Value of $\sec \beta$ .

Answer to Problem 42E

The value of $\sec \beta$ is $\frac{4}{\sqrt{7}}$

Explanation of Solution

Given info:

  $\cos \beta =\frac{\sqrt{7}}{4}$

Formula used:

We have the results

  $\begin{array}{l}\sin \theta =\cos (90-\theta )\\ \cos \theta =\sin (90-\theta )\\ \cos ec\theta =\sec (90-\theta )\\ \sec \theta =\cos ec(90-\theta )\\ \tan \theta =\cot (90-\theta )\\ \cot \theta =\tan (90-\theta )\\ tan\theta =\frac{\sin \theta }{\cos \theta }\\ \cot \theta =\frac{\cos \theta }{\sin \theta }\\ cosec\theta =\frac{1}{\sin \theta }\\ \sec \theta =\frac{1}{\cos \theta }\\ si{n}^2\theta +{\cos }^2\theta =1\\ {\sec }^2\theta -{\tan }^2\theta =1\\ cose{c}^2\theta -{\cot }^2\theta =1\end{array}$

Calculation:

Here we need to find $\sec \beta$ using $\cos \beta =\frac{\sqrt{7}}{4}$

We have

  $\begin{array}{l}\sec \beta =\frac{1}{\cos \beta }\\ \sec \beta =\frac{1}{\frac{\sqrt{7}}{4}}=\frac{4}{\sqrt{7}}\end{array}$

Value of $\sec \beta$ is $\frac{4}{\sqrt{7}}$

Conclusion:

Thus,the value of $\sec \beta$ is $\frac{4}{\sqrt{7}}$

(b)

To determine

Value of $\sin \beta$ .

Answer to Problem 42E

The value of $\sin \beta$ is $\frac{3}{4}$

Explanation of Solution

Given info:

  $\cos \beta =\frac{\sqrt{7}}{4}$

Formula used:

We have the results

  $\begin{array}{l}\sin \theta =\cos (90-\theta )\\ \cos \theta =\sin (90-\theta )\\ \cos ec\theta =\sec (90-\theta )\\ \sec \theta =\cos ec(90-\theta )\\ \tan \theta =\cot (90-\theta )\\ \cot \theta =\tan (90-\theta )\\ tan\theta =\frac{\sin \theta }{\cos \theta }\\ \cot \theta =\frac{\cos \theta }{\sin \theta }\\ cosec\theta =\frac{1}{\sin \theta }\\ \sec \theta =\frac{1}{\cos \theta }\\ si{n}^2\theta +{\cos }^2\theta =1\\ {\sec }^2\theta -{\tan }^2\theta =1\\ cose{c}^2\theta -{\cot }^2\theta =1\end{array}$

Calculation:

Here we need to find $\sin \beta$ using $\cos \beta =\frac{\sqrt{7}}{4}$

We have

  $\begin{array}{l}si{n}^2\beta +{\cos }^2\beta =1\\ si{n}^2\beta +\frac{7}{16}=1\\ sin\beta =\frac{3}{4}\end{array}$

Value of $\sin \beta$ is $\frac{3}{4}$

Conclusion:

Thus,the value of $\sin \beta$ is $\frac{3}{4}$

(c)

To determine

Value of $\cot \beta$ .

Answer to Problem 42E

The value of $\cot \beta$ is $\frac{\sqrt{7}}{3}$

Explanation of Solution

Given info:

  $\cos \beta =\frac{\sqrt{7}}{4}$

Formula used:

We have the results

  $\begin{array}{l}\sin \theta =\cos (90-\theta )\\ \cos \theta =\sin (90-\theta )\\ \cos ec\theta =\sec (90-\theta )\\ \sec \theta =\cos ec(90-\theta )\\ \tan \theta =\cot (90-\theta )\\ \cot \theta =\tan (90-\theta )\\ tan\theta =\frac{\sin \theta }{\cos \theta }\\ \cot \theta =\frac{\cos \theta }{\sin \theta }\\ cosec\theta =\frac{1}{\sin \theta }\\ \sec \theta =\frac{1}{\cos \theta }\\ si{n}^2\theta +{\cos }^2\theta =1\\ {\sec }^2\theta -{\tan }^2\theta =1\\ cose{c}^2\theta -{\cot }^2\theta =1\end{array}$

Calculation:

Here we need to find $\cot \beta$ using $\cos \beta =\frac{\sqrt{7}}{4}$

From part(b) we have the value of $\sin \beta$ is $\frac{3}{4}$

We have

  $\begin{array}{l}\cot \beta =\frac{\cos \beta }{\sin \beta }\\ \cot \beta =\frac{\frac{\sqrt{7}}{4}}{\frac{3}{4}}=\frac{\sqrt{7}}{3}\end{array}$

Value of $\cot \beta$ is $\frac{\sqrt{7}}{3}$

Conclusion:

Thus,the value of $\cot \beta$ is $\frac{\sqrt{7}}{3}$

(d)

To determine

Value of $\sin (90-\beta )$ .

Answer to Problem 42E

The value of $\sin (90-\beta )$ is $\frac{\sqrt{7}}{4}$

Explanation of Solution

Given info:

  $\cos \beta =\frac{\sqrt{7}}{4}$

Formula used:

We have the results

  $\begin{array}{l}\sin \theta =\cos (90-\theta )\\ \cos \theta =\sin (90-\theta )\\ \cos ec\theta =\sec (90-\theta )\\ \sec \theta =\cos ec(90-\theta )\\ \tan \theta =\cot (90-\theta )\\ \cot \theta =\tan (90-\theta )\\ tan\theta =\frac{\sin \theta }{\cos \theta }\\ \cot \theta =\frac{\cos \theta }{\sin \theta }\\ cosec\theta =\frac{1}{\sin \theta }\\ \sec \theta =\frac{1}{\cos \theta }\\ si{n}^2\theta +{\cos }^2\theta =1\\ {\sec }^2\theta -{\tan }^2\theta =1\\ cose{c}^2\theta -{\cot }^2\theta =1\end{array}$

Calculation:

Here we need to find $\sin (90-\beta )$ using $\cos \beta =\frac{\sqrt{7}}{4}$

We have

  $\begin{array}{l}\sin (90-\beta )=\cos \beta \\ \sin (90-\beta )=\frac{\sqrt{7}}{4}\end{array}$

Value of $\sin (90-\beta )$ is $\frac{\sqrt{7}}{4}$

Conclusion:

Thus,the value of $\sin (90-\beta )$ is $\frac{\sqrt{7}}{4}$