Chapter 4.3, Problem 55E

a)

To determine

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

Answer to Problem 55E

  $\tan \alpha =\frac{1}{4}$

Explanation of Solution

Given: $\cot \alpha =4$

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

Calculation: we have $\cot \alpha =4$

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

  $\tan \alpha =\frac{1}{4}$ (Putting value of $\cot \alpha =4$ )

Conclusion: $\tan \alpha =\frac{1}{4}$

b)

To determine

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

Answer to Problem 55E

  $\csc \alpha =\sqrt{17}$

Explanation of Solution

Given: $\cot \alpha =4$

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

Calculation: We have $\cot \alpha =4$

  $1+{\cot }^2\alpha ={\csc }^2\alpha$ (Pythagorean identity)$1+{\left(4\right)}^2={\csc }^2\alpha$

  $17={\csc }^2\alpha$ (Simplify)

  $\csc \alpha =\sqrt{17}$

Conclusion: $\csc \alpha =\sqrt{17}$

c)

To determine

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

Answer to Problem 55E

  $\sec \alpha =\frac{\sqrt{17}}{4}$

Explanation of Solution

Given: $\cot \alpha =4$

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

Calculation: We have $\cot \alpha =4$

Since, $\tan \alpha =\frac{1}{4}$ (As proved in (a) part)

  $1+{\left(\frac{1}{4}\right)}^2={\sec }^2\alpha$ Using $1+{\tan }^2\alpha ={\sec }^2\alpha$ (Pythagorean identity)

  $1+\frac{1}{16}={\sec }^2\alpha$ (Simplify)

  ${\sec }^2\alpha =\frac{17}{16}$ (Simplify)

  $\sec \alpha =\frac{\sqrt{17}}{4}$ (Taking square root)

Conclusion: $\sec \alpha =\frac{\sqrt{17}}{4}$

d)

To determine

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

Answer to Problem 55E

  $\tan \left(90-\alpha \right)=4$

Explanation of Solution

Given: $\cot \alpha =4$

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

Calculation: We have $\cot \alpha =4$

  $\tan \left(90-\alpha \right)=4$ ( $\tan \left(90-\alpha \right)=\cot \alpha$ )

Conclusion: $\tan \left(90-\alpha \right)=4$