You must answer in the text field below for Parts 1 and 2, identifying each Part clearly for point-credit assignment. Pager submission is REQUIRED for Part 3.
Given the polar equation, answer the following. Use T for e.
sec(20)
Part 1. What is the first step to rewrite this equation in rectangular form?
Part 2. What is the second step?
Part 3. From the choices below, select the letter of the rectangular equation. Show solution on paper.
A. y = 2x
B. 2x - y2 = 0
C. y = 2/x
D. y - 2x2 = 0
E. y= - 2x
F. not here
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12pt v Paragraph y

**Part 1: What is the first step to rewrite this equation in rectangular form?**

The first step to rewrite a polar equation in rectangular form is to express the trigonometric function in terms of \( r \) and \( \theta \). In this case, for \( \sec(2\theta) \), we use the identity \( \sec(2\theta) = \frac{1}{\cos(2\theta)} \).

**Part 2: What is the second step?**

The second step is to use the double-angle formula for cosine to convert the expression involving \( \theta \) to one involving \( r \) and \( \theta \). The double-angle formula for cosine is \( \cos(2\theta) = 2\cos^2(\theta) - 1 \). Then, replace \( \cos(\theta) \) and \( \sin(\theta) \) with their rectangular coordinates equivalents \( \cos(\theta) = \frac{x}{r} \) and \( \sin(\theta) = \frac{y}{r} \).

**Part 3:**

To complete Part 3, I'll derive the solution step-by-step:

Given \( \sec(2\theta) = T \):

1. Rewrite using the secant identity:
   \[
   \sec(2\theta) = \frac{1}{\cos(2\theta)}
   \]
   Therefore,
   \[
   \frac{1}{\cos(2\theta)} = T
   \]
   which gives us:
   \[
   \cos(2\theta) = \frac{1}{T}
   \]

2. Use the double-angle formula for cosine:
   \[
   \cos(2\theta) = 2\cos^2(\theta) - 1
   \]
   Substitute \( \cos(2\theta) \) in the equation:
   \[
   2\cos^2(\theta) - 1 = \frac{1}{T}
   \]

3. Express \( \cos(\theta) \) in terms of \( x \) and \( r \):
   \[
   \cos(\theta) = \frac{x}{r}
   \]
   Therefore,
   \[
   2\left(\frac{x}{r}\right)^2 - 1 = \frac{1}{T}
   \]

4. Simplify the equation:
   \[
   2\frac{x^2}{r^2} - 1 = \frac{1}{T}
   \]
   \[
   2\frac{x^2}{x^2 + y^2} - 1 = \frac{1}{T}
   \]
   Multiply through by \( (x^2 + y^2) \):
   \[
   2x^2 - (x^2 + y^2) = \frac{x^2 + y^2}{T}
   \]
   \[
   2x^2 - x^2 - y^2 = \frac{x^2 + y^2}{T}
   \]
   \[
   x^2 - y^2 = \frac{x^2 + y^2}{T}
   \]

5. Rearrange to find a solution:
   \[
   Tx^2 - Ty^2 = x^2 + y^2
   \]
   \[
   Tx^2 - x^2 = Ty^2 + y^2
   \]
   \[
   (T - 1)x^2 = (T + 1)y^2
   \]

Therefore, the rectangular form of the equation is:

\[
y = \pm \sqrt{\frac{T-1}{T+1}}x
\]

From the choices given, this does not directly match any of the options A-F. So the correct answer is:

**F. not here**

Do you want further details or have any questions?

Here are some relative questions that expand on the previous information:

1. What are the key differences between polar and rectangular coordinates?
2. How do you convert from polar to rectangular coordinates?
3. What are some practical applications of polar coordinates?
4. How do you use the Pythagorean identity in polar coordinates?
5. Can all polar equations be converted to rectangular form?
6. What is the significance of the secant function in trigonometry?
7. How can we use double-angle formulas in solving trigonometric equations?
8. What are some common pitfalls in converting polar equations to rectangular form?

**Tip:** Always remember to express trigonometric functions in terms of \( x \) and \( y \) using the fundamental relationships between polar and rectangular coordinates.

Learn how to convert a polar equation involving sec(2θ) to rectangular form step-by-step. This problem covers trigonometric identities, secant functions, and the use of double-angle formulas for cosine. Suitable for advanced high school students and beyond.

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