Answered

Welcome to Westonci.ca, where you can find answers to all your questions from a community of experienced professionals. Get detailed answers to your questions from a community of experts dedicated to providing accurate information. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______
a) sec sqr Theta
b) Cot theta
c) cot (pie/2-Theta)
d) csc sqr theta

Sagot :

MrRoyal

Answer:

[tex](a)\ \sec^2(\theta) = 82[/tex]

[tex](b)\ \cot(\theta) = \frac{1}{9}[/tex]

[tex](c)\ \cot(\frac{\pi}{2} - \theta) = 9[/tex]

[tex](d)\ \csc^2(\theta) = \frac{82}{81}[/tex]

Step-by-step explanation:

Given

[tex]\tan(\theta) = 9[/tex]

Required

Solve (a) to (d)

Using tan formula, we have:

[tex]\tan(\theta) = \frac{Opposite}{Adjacent}[/tex]

This gives:

[tex]\frac{Opposite}{Adjacent} = 9[/tex]

Rewrite as:

[tex]\frac{Opposite}{Adjacent} = \frac{9}{1}[/tex]

Using a unit ratio;

[tex]Opposite = 9; Adjacent = 1[/tex]

Using Pythagoras theorem, we have:

[tex]Hypotenuse^2 = Opposite^2 + Adjacent^2[/tex]

[tex]Hypotenuse^2 = 9^2 + 1^2[/tex]

[tex]Hypotenuse^2 = 81 + 1[/tex]

[tex]Hypotenuse^2 = 82[/tex]

Take square roots of both sides

[tex]Hypotenuse =\sqrt{82}[/tex]

So, we have:

[tex]Opposite = 9; Adjacent = 1[/tex]

[tex]Hypotenuse =\sqrt{82}[/tex]

Solving (a):

[tex]\sec^2(\theta)[/tex]

This is calculated as:

[tex]\sec^2(\theta) = (\sec(\theta))^2[/tex]

[tex]\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2[/tex]

Where:

[tex]\cos(\theta) = \frac{Adjacent}{Hypotenuse}[/tex]

[tex]\cos(\theta) = \frac{1}{\sqrt{82}}[/tex]

So:

[tex]\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2[/tex]

[tex]\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2[/tex]

[tex]\sec^2(\theta) = (\sqrt{82})^2[/tex]

[tex]\sec^2(\theta) = 82[/tex]

Solving (b):

[tex]\cot(\theta)[/tex]

This is calculated as:

[tex]\cot(\theta) = \frac{1}{\tan(\theta)}[/tex]

Where:

[tex]\tan(\theta) = 9[/tex] ---- given

So:

[tex]\cot(\theta) = \frac{1}{\tan(\theta)}[/tex]

[tex]\cot(\theta) = \frac{1}{9}[/tex]

Solving (c):

[tex]\cot(\frac{\pi}{2} - \theta)[/tex]

In trigonometry:

[tex]\cot(\frac{\pi}{2} - \theta) = \tan(\theta)[/tex]

Hence:

[tex]\cot(\frac{\pi}{2} - \theta) = 9[/tex]

Solving (d):

[tex]\csc^2(\theta)[/tex]

This is calculated as:

[tex]\csc^2(\theta) = (\csc(\theta))^2[/tex]

[tex]\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2[/tex]

Where:

[tex]\sin(\theta) = \frac{Opposite}{Hypotenuse}[/tex]

[tex]\sin(\theta) = \frac{9}{\sqrt{82}}[/tex]

So:

[tex]\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2[/tex]

[tex]\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2[/tex]

[tex]\csc^2(\theta) = \frac{82}{81}[/tex]

We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.