Chapter 5.1, Problem 2E

(a)

To determine

To match the function ${\sin }^{2}u$ with an equivalent expression.

Answer to Problem 2E

  $\left(ii\right)1-{\cos }^{2}u$

Explanation of Solution

Given:

The expressions,

  $\begin{array}{l}\left(i\right)1+{\cot }^{2}u\\ \left(ii\right)1-{\cos }^{2}u\\ \left(iii\right)1+{\tan }^{2}u\end{array}$

Concept Used:

Pythagorean Identity:

  ${\sin }^2\theta +{\cos }^2\theta =1$

Calculation:

In order tofind the equivalent expression for the function ${\sin }^{2}u$ , use the Pythagorean identity and simplify further as shown below:

  $\begin{array}{c}{\sin }^{2}u+{\cos }^{2}u=1\\ \Rightarrow {\sin }^{2}u=1-{\cos }^{2}u\end{array}$

Thus, the equivalent expression to ${\sin }^{2}u$ is $1-{\cos }^{2}u$ .

(b)

To determine

To match the function ${\sec }^{2}u$ with an equivalent expression.

Answer to Problem 2E

  $\left(iii\right)1+{\tan }^{2}u$

Explanation of Solution

Given:

The expressions,

  $\begin{array}{l}\left(i\right)1+{\cot }^{2}u\\ \left(ii\right)1-{\cos }^{2}u\\ \left(iii\right)1+{\tan }^{2}u\end{array}$

Concept Used:

Pythagorean Identity:

  ${\sin }^2\theta +{\cos }^2\theta =1$

Reciprocal identities of trigonometric functions:

  $\begin{array}{l}\sin \theta =\frac{1}{\csc \theta },\csc \theta =\frac{1}{\sin \theta },\\ \cos \theta =\frac{1}{\sec \theta },\sec \theta =\frac{1}{\cos \theta },\\ \tan \theta =\frac{1}{\cot \theta },\cot \theta =\frac{1}{\tan \theta }\end{array}$

Calculation:

In order to find the equivalent expression for the function ${\sec }^{2}u$ , use the Pythagorean identity and simplify further by dividing both sides by ${\cos }^{2}u$ as shown below:

  $\begin{array}{c}{\sin }^{2}u+{\cos }^{2}u=1\\ \frac{{\sin }^{2}u}{{\cos }^{2}u}+\frac{{\cos }^{2}u}{{\cos }^{2}u}=\frac{1}{{\cos }^{2}u}\\ {\tan }^{2}u+1={\sec }^{2}u\\ \Rightarrow {\sec }^{2}u=1+{\tan }^{2}u\end{array}$

Thus, the equivalent expression to ${\sec }^{2}u$ is $1+{\tan }^{2}u$ .

(c)

To determine

To match the function ${\csc }^{2}u$ with an equivalent expression.

Answer to Problem 2E

  $\left(i\right)1+{\cot }^{2}u$

Explanation of Solution

Given:

The expressions,

  $\begin{array}{l}\left(i\right)1+{\cot }^{2}u\\ \left(ii\right)1-{\cos }^{2}u\\ \left(iii\right)1+{\tan }^{2}u\end{array}$

Concept Used:

Pythagorean Identity:

  ${\sin }^2\theta +{\cos }^2\theta =1$

Reciprocal identities of trigonometric functions:

  $\begin{array}{l}\sin \theta =\frac{1}{\csc \theta },\csc \theta =\frac{1}{\sin \theta },\\ \cos \theta =\frac{1}{\sec \theta },\sec \theta =\frac{1}{\cos \theta },\\ \tan \theta =\frac{1}{\cot \theta },\cot \theta =\frac{1}{\tan \theta }\end{array}$

Calculation:

In order to find the equivalent expression for the function ${\csc }^{2}u$ , use the Pythagorean identity and simplify further by dividing both sides by ${\sin }^{2}u$ as shown below:

  $\begin{array}{c}{\sin }^{2}u+{\cos }^{2}u=1\\ \frac{{\sin }^{2}u}{{\sin }^{2}u}+\frac{{\cos }^{2}u}{{\sin }^{2}u}=\frac{1}{{\sin }^{2}u}\\ 1+{\cot }^{2}u={\csc }^{2}u\\ \Rightarrow {\csc }^{2}u=1+{\cot }^{2}u\end{array}$

Thus, the equivalent expression to ${\csc }^{2}u$ is $1+{\cot }^{2}u$ .