Chapter 4.3, Problem 54E

a)

To determine

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

Answer to Problem 54E

The exact value is $\cos \theta =\frac{1}{5}$.

Explanation of Solution

Given: $\sec \theta =5,\tan \theta =2\sqrt{6}$.

Concept Used: Using reciprocal identity $\cos \theta =\frac{1}{\sec \theta }$

Calculation: We have $\sec \theta =5$.

By using reciprocal identity, $\cos \theta =\frac{1}{\sec \theta }$.

  $\cos \theta =\frac{1}{5}$.(Putting value of $\sec \theta =5$ )

Conclusion: $\cos \theta =\frac{1}{5}$.

b)

To determine

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

Answer to Problem 54E

The exact value is $\cot \theta =\frac{\sqrt{6}}{12}$.

Explanation of Solution

Given: $\sec \theta =5,\tan \theta =2\sqrt{6}$

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

Calculation: We have $\tan \theta =2\sqrt{6}$

  $\cot \theta =\frac{1}{2\sqrt{6}}$ (Putting the value of $\tan \theta =2\sqrt{6}$)$\cot \theta =\frac{1}{2\sqrt{6}}.\left(\frac{\sqrt{6}}{\sqrt{6}}\right)=\frac{\sqrt{6}}{12}$ (Multiply and divide by $\sqrt{6}$ )

  $\cot \theta =\frac{\sqrt{6}}{12}$

Conclusion: The exact value is $\cot \theta =\frac{\sqrt{6}}{12}$.

c)

To determine

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

Answer to Problem 54E

  $\cot \left(90-\theta \right)=2\sqrt{6}$

Explanation of Solution

Given: $\sec \theta =5,\tan \theta =2\sqrt{6}$

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

Calculation: We have $\tan \theta =2\sqrt{6}$

  $\cot \left(90-\theta \right)=2\sqrt{6}$ ( $\cot \left(90-\theta \right)=\tan \theta$ )

Conclusion: $\cot \left(90-\theta \right)=2\sqrt{6}$

d)

To determine

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

Answer to Problem 54E

The exact value is $\sin \theta =\frac{2\sqrt{6}}{5}$ .

Explanation of Solution

Given: $\sec \theta =5,\tan \theta =2\sqrt{6}$

Concept Use: Using Pythagorean identity ${\sin }^2\theta +{\cos }^2\theta =1$

Calculation: We have $\cos \theta =\frac{1}{5}$ (As proved in above part)

  ${\sin }^2\theta +{\cos }^2\theta =1$ (Pythagorean identity)

  ${\sin }^2\theta +{\left(\frac{1}{5}\right)}^2=1$ (Putting the value $\cos \theta =\frac{1}{5}$ )

  ${\sin }^2\theta =1-\frac{1}{25}$ (Simplify)

  ${\sin }^2\theta =\frac{24}{25}$ (Simplify)

  $\sin \theta =\frac{2\sqrt{6}}{5}$ (Taking square root)

Conclusion: $\sin \theta =\frac{2\sqrt{6}}{5}$