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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]