Chapter 4.3, Problem 40E

(a)

To determine

Value of $\cos \theta$ .

Answer to Problem 40E

The value of $\cos \theta$ is $\frac{1}{5}$

Explanation of Solution

Given info:

  $\cos \theta =\frac{1}{3}$

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 $\cos \theta$ using $\sec \theta =5$

We have

  $\begin{array}{l}\sec \theta =\frac{1}{\cos \theta }\\ \cos \theta =\frac{1}{\sec \theta }\\ \cos \theta =\frac{1}{5}\end{array}$

Value of $\cos \theta$ is $\frac{1}{5}$

Conclusion:

Thus,the value of $\cos \theta$ is $\frac{1}{5}$

(b)

To determine

Value of $\cot \theta$ .

Answer to Problem 40E

The value of $\cot \theta$ is $\frac{1}{\sqrt{24}}$

Explanation of Solution

Given info:

  $\cos \theta =\frac{1}{3}$

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 $\cot \theta$ using $\sec \theta =5$

We have

  $\begin{array}{l}{\sec }^2\theta -{\tan }^2\theta =1\\ 25-{\tan }^2\theta =1\\ \tan \theta =\sqrt{24}\\ \tan \theta =\frac{1}{\cot \theta }=\sqrt{24}\\ \cot \theta =\frac{1}{\sqrt{24}}\end{array}$

Value of $\cot \theta$ is $\frac{1}{\sqrt{24}}$

Conclusion:

Thus,the value of $\cot \theta$ is $\frac{1}{\sqrt{24}}$

(c)

To determine

Value of $\cot (90-\theta )$ .

Answer to Problem 40E

The value of $\cot (90-\theta )$ is $\sqrt{24}$

Explanation of Solution

Given info:

  $\cos \theta =\frac{1}{3}$

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 $\cot (90-\theta )$ using $\sec \theta =5$

We have

  $\begin{array}{l}{\sec }^2\theta -{\tan }^2\theta =1\\ 25-{\tan }^2\theta =1\\ \tan \theta =\sqrt{24}\\ \tan \theta =\cot (90-\theta )=\sqrt{24}\\ \cot (90-\theta )=\sqrt{24}\end{array}$

Value of $\cot (90-\theta )$ is $\sqrt{24}$

Conclusion:

Thus,the value of $\cot (90-\theta )$ is $\sqrt{24}$

(d)

To determine

Value of $\sin \theta$ .

Answer to Problem 40E

The value of $\sin \theta$ is $\frac{\sqrt{24}}{5}$

Explanation of Solution

Given info:

  $\cos \theta =\frac{1}{3}$

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 $\sin \theta$ using $\sec \theta =5$

We have

  $\begin{array}{l}\sec \theta =5\\ \cos \theta =\frac{1}{5}\\ si{n}^2\theta +{\cos }^2\theta =1\\ si{n}^2\theta +\frac{1}{25}=1\\ sin\theta =\frac{\sqrt{24}}{5}\end{array}$

Value of $\sin \theta$ is $\frac{\sqrt{24}}{5}$

Conclusion:

Thus,the value of $\sin \theta$ is $\frac{\sqrt{24}}{5}$