Chapter 4.3, Problem 56E

a)

To determine

To calculate:The exact value of each indicated trigonometric function by using the given function.

Answer to Problem 56E

  $\cot \beta =\frac{1}{3}$

Explanation of Solution

Given: $\tan \beta =3$

Concept Used: Using reciprocal identity $\cot \beta =\frac{1}{\tan \beta }$

Calculation: we have $\tan \beta =3$

By using reciprocal identity $\cot \beta =\frac{1}{\tan \beta }$

  $\cot \beta =\frac{1}{3}$ (Putting value of $\tan \beta =3$ )

Conclusion: $\cot \beta =\frac{1}{3}$

b)

To determine

To calculate:The exact value of each indicated trigonometric function by using the given function

Answer to Problem 56E

  $\cos \beta =\frac{1}{\sqrt{10}}$

Explanation of Solution

Given: $\tan \beta =3$

Concept Used: $1+{\tan }^2\beta ={\sec }^2\beta$ (Pythagorean identity)

  $\cos \beta =\frac{1}{\sec \beta }$ (Reciprocal identity)

Calculation:We have $\tan \beta =3$

  $1+{\tan }^2\beta ={\sec }^2\beta$ (Pythagorean identity)

  $1+{\left(3\right)}^2={\sec }^2\beta$

  $10={\sec }^2\beta$ (Simplify)

  $\sec \beta =\sqrt{10}$

  $\cos \beta =\frac{1}{\sec \beta }$ (Reciprocal identity)

  $\cos \beta =\frac{1}{\sqrt{10}}$

Conclusion: $\cos \beta =\frac{1}{\sqrt{10}}$

c)

To determine

To calculate:The exact value of each indicated trigonometric function by using the given function.

Answer to Problem 56E

Explanation of Solution

Given: $\tan \beta =3$

Concept Used: Using Trigonometric ratios $\tan \left(90-\beta \right)=\cot \beta$ (co functions of complimentary angles are equal).

Calculation: We have $\tan \beta =3$

By using Trigonometric ratios $\tan \left(90-\beta \right)=\cot \beta$

Since, $\cot \beta =\frac{1}{3}$ (As proved in above (a) part)

  $\Rightarrow \tan \left(90-\beta \right)=\frac{1}{3}$

Conclusion: $\tan \left(90-\beta \right)=\frac{1}{3}$

d)

To determine

To calculate:The exact value of each indicated trigonometric function by using the given function

Answer to Problem 56E

  $\csc \beta =\frac{\sqrt{10}}{3}$

Explanation of Solution

Given: $\tan \beta =3$

Concept Used: $1+{\cot }^2\beta ={\csc }^2\beta$ (Pythagorean identity)

Calculation: We have $\tan \beta =3$

  $1+{\cot }^2\beta ={\csc }^2\beta$ (Pythagorean identity)

  $1+{\left(\frac{1}{3}\right)}^2={\csc }^2\beta$ ( $\cot \beta =\frac{1}{3}$ )(As proved in above (a) part)

  $1+\frac{1}{9}={\csc }^2\beta$ (Simplify)

  ${\csc }^2\beta =\frac{10}{9}$ (Simplify)

  $\csc \beta =\frac{\sqrt{10}}{3}$

  $$ (Taking square root)

Conclusion: $\csc \beta =\frac{\sqrt{10}}{3}$