Chapter 4.3, Problem 38E

(a)

To determine

Value of $\csc {30}^0$ .

Answer to Problem 38E

The value of $\csc {30}^0$ is $2$

Explanation of Solution

Given info:

  $\begin{array}{l}\sin {30}^0=\frac{1}{2}\\ \tan {30}^0=\frac{\sqrt{3}}{3}\end{array}$

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 $\csc {30}^0$ using $\sin {30}^0=\frac{1}{2},\tan {30}^0=\frac{\sqrt{3}}{3}$

We have

  $\begin{array}{l}cosec\theta =\frac{1}{\sin \theta }\\ cosec30=\frac{1}{\sin 30}\\ cosec30=\frac{1}{\frac{1}{2}}=2\end{array}$

Value of $\csc {30}^0=2$

Conclusion:

Thus,the value of $\csc {30}^0$ is $2$

(b)

To determine

Value of $\cot {60}^0$ .

Answer to Problem 38E

The value of $\cot {60}^0$ is $\frac{\sqrt{3}}{3}$

Explanation of Solution

Given info:

  $\begin{array}{l}\sin {30}^0=\frac{1}{2}\\ \tan {30}^0=\frac{\sqrt{3}}{3}\end{array}$

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 $\cot {60}^0$ using $\sin {30}^0=\frac{1}{2},\tan {30}^0=\frac{\sqrt{3}}{3}$

We have

  $\begin{array}{l}\cot \theta =\tan (90-\theta )\\ \cot 60=\tan (90-60)\\ \cot 60=\tan 30=\frac{\sqrt{3}}{3}\end{array}$

Value of $\cot {60}^0$ is $\frac{\sqrt{3}}{3}$

Conclusion:

Thus,the value of $\cot {60}^0$ is $\frac{\sqrt{3}}{3}$

(c)

To determine

Value of $\cos {30}^0$ .

Answer to Problem 38E

The value of $\cos {30}^0$ is $\frac{\sqrt{3}}{2}$

Explanation of Solution

Given info:

  $\begin{array}{l}\sin {30}^0=\frac{1}{2}\\ \tan {30}^0=\frac{\sqrt{3}}{3}\end{array}$

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 $\cos {30}^0$ using $\sin {30}^0=\frac{1}{2},\tan {30}^0=\frac{\sqrt{3}}{3}$

We have

  $\begin{array}{l}si{n}^2\theta +{\cos }^2\theta =1\\ si{n}^{2}30+{\cos }^{2}30=1\\ \frac{1}{4}+{\cos }^{2}30=1\\ \cos 30=\frac{\sqrt{3}}{2}\end{array}$

Value of $\cos {30}^0$ is $\frac{\sqrt{3}}{2}$

Conclusion:

Thus,the value of $\cos {30}^0$ is $\frac{\sqrt{3}}{2}$

(d)

To determine

Value of $\cot {30}^0$ .

Answer to Problem 38E

The value of $\cot {30}^0$ is $\sqrt{3}$

Explanation of Solution

Given info:

  $\begin{array}{l}\sin {30}^0=\frac{1}{2}\\ \tan {30}^0=\frac{\sqrt{3}}{3}\end{array}$

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 $\cot {30}^0$ using $\sin {30}^0=\frac{1}{2},\tan {30}^0=\frac{\sqrt{3}}{3}$

We have

  $\begin{array}{l}\cot \theta =\frac{1}{\tan \theta }\\ \cot 30=\frac{1}{\tan 30}=\frac{3}{\sqrt{3}}=\sqrt{3}\end{array}$

Value of $\cot {30}^0$ is $\sqrt{3}$

Conclusion:

Thus,the value of $\cot {30}^0$ is $\sqrt{3}$