Pre-Calculus Assignment: Trigonometry Problem Solutions and Answers

This pre-calculus assignment solution addresses several trigonometry problems. The document presents step-by-step solutions for various trigonometric functions and identities, including sine, cosine, tangent, cotangent, secant, and cosecant. The solutions cover the application of trigonometric formulas and identities to find unknown values and relationships. The assignment references key concepts such as the Pythagorean identity and the relationship between trigonometric functions. The solution also includes a bibliography of sources used in the assignment, providing a comprehensive overview of the problem-solving process and the underlying mathematical principles. This resource is designed to help students understand and solve similar pre-calculus problems.

Running head: PRE-CALCULUS
Pre-Calculus
Name of the student:
Name of the University:
Author note:

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PRE-CALCULUS
Table of Contents
Answer to the question 25..........................................................................................................1
Answer to the question 26..........................................................................................................1
Answer to the question 27..........................................................................................................1
Answer to the question 28..........................................................................................................2
Answer to the question 29..........................................................................................................3
Answer to the question 30..........................................................................................................3
Bibliography...............................................................................................................................5
Answer to the question 25

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PRE-CALCULUS
Given that
tanθ =3/4
cot θ =4/3
Now
1+ cot2θ= csc2
θ
Or 1+ (4/3)2= csc2θ
Or csc2
θ= 25/9
Or cscθ= 5/3
Answer to the question 26
Given that
S ec θ = 2
Now
1+ tan2θ= sec2θ
Or tan2
θ= 4-1
Or 1/ cot2θ= 3
Or cot θ= 1
√ 3
Answer to the question 27
Given that
sin θ = 4/5
cos θ= sqrt (1- sin2θ)

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PRE-CALCULUS
= sqrt (1-(4/5)2)
= Sqrt (1-16/25)
= 3/5
Hence tan θ= sin θ
cos θ
=
4
5
3
5
tan θ = 4/3
Answer to the question 28
Given that
Sec θ = 5/4
Now 1+ tan2
θ= sec2
θ
Or tan2
θ= (5/4)2-1
= 9/16
tan θ =3/4
cot θ =4/3
Answer to the question 29
Given that
S∈θ = 3 √13
13
Then

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PRE-CALCULUS
sin2
θ = 9∗13
13∗13
= 9/13
Now
cos θ= sqrt (1-sin2θ)
= sqrt (1-9/13)
= sqrt (4/13)
cos θ= 2
√13
Therefore
sec θ= √ 13
2
Answer to the question 30
Given that
sin θ = 12
13
Then
sin2
θ = 144
169
cos θ= sqrt (1-sin2θ)
= sqrt (1-144/168)
= sqrt (25/169)
cos θ= 5/13

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PRE-CALCULUS
Hence
cot θ=cos θ
sinθ
=
5
13
12
13
=5/12
Bibliography
Larson, R., 2013. Algebra & Trigonometry. Cengage Learning.

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PRE-CALCULUS
Wildberger, N.J., 2013. Universal hyperbolic geometry I: trigonometry. Geometriae
Dedicata, 163(1), pp.215-274.