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

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]

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