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The system of equations has infinitely many solutions, meaning there are multiple values of \( A \) that satisfy the given relationships.
Solución
To solve the given system of equations, we have:
1. \( C = \sec A + \tan A \)
2. \( a = \sec A - \tan A \)
3. \( C = \frac{\csc A + 1}{\csc A - 1} \)
We will solve these equations step by step.
### Step 1: Express \( C \) in terms of \( A \)
From the first equation, we have:
\[
C = \sec A + \tan A
\]
### Step 2: Express \( a \) in terms of \( A \)
From the second equation, we have:
\[
a = \sec A - \tan A
\]
### Step 3: Relate \( C \) and \( a \)
Now, we can express \( C \) and \( a \) in terms of \( \sec A \) and \( \tan A \):
- Adding the two equations:
\[
C + a = (\sec A + \tan A) + (\sec A - \tan A) = 2\sec A
\]
- Subtracting the second equation from the first:
\[
C - a = (\sec A + \tan A) - (\sec A - \tan A) = 2\tan A
\]
### Step 4: Solve for \( \sec A \) and \( \tan A \)
From the equations \( C + a = 2\sec A \) and \( C - a = 2\tan A \), we can express \( \sec A \) and \( \tan A \) as:
\[
\sec A = \frac{C + a}{2}
\]
\[
\tan A = \frac{C - a}{2}
\]
### Step 5: Substitute into the third equation
Now we substitute \( C \) into the third equation:
\[
C = \frac{\csc A + 1}{\csc A - 1}
\]
### Step 6: Express \( \csc A \) in terms of \( C \)
We can rearrange the third equation to express \( \csc A \):
\[
C(\csc A - 1) = \csc A + 1
\]
\[
C\csc A - C = \csc A + 1
\]
\[
C\csc A - \csc A = C + 1
\]
\[
\csc A(C - 1) = C + 1
\]
\[
\csc A = \frac{C + 1}{C - 1}
\]
### Step 7: Solve the system of equations
Now we have:
1. \( C = \sec A + \tan A \)
2. \( a = \sec A - \tan A \)
3. \( \csc A = \frac{C + 1}{C - 1} \)
We can now solve the system of equations. Let's proceed to solve these equations together.
I'll use the tool to solve the system of equations.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
\(\left\{ \begin{array}{l}C=\frac{C+1}{C-1}+\frac{C-a}{2}\\a=\frac{C+a}{2}-\frac{C-a}{2}\end{array}\right.\)
- step1: Calculate:
\(\left\{ \begin{array}{l}C=\frac{C+1}{C-1}+\frac{C-a}{2}\\2a=2a\end{array}\right.\)
- step2: Solve the equation:
\(a=\frac{C^{2}-3C-2}{-C+1}\)
- step3: Calculate:
\(\left(C,a\right) = \left(C,\frac{C^{2}-3C-2}{-C+1}\right),C \in \mathbb{R}\)
- step4: Alternative Form:
\(\textrm{Infinitely many solutions}\)
The system of equations has infinitely many solutions. This indicates that the equations are dependent, meaning they describe the same relationship in different forms.
### Summary of Findings:
1. We derived expressions for \( C \) and \( a \) in terms of \( \sec A \) and \( \tan A \).
2. We also expressed \( \csc A \) in terms of \( C \).
3. The relationships among these variables lead to a consistent set of equations that do not yield a unique solution but rather a family of solutions.
If you have specific values for \( a \) or \( C \), we can further explore particular solutions. Otherwise, the relationships established can be used to find values of \( A \) that satisfy the equations for any chosen \( C \) or \( a \).
Revisado y aprobado por el equipo de tutoría de UpStudy
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