If `A=C_(0) -C_(2) +C_(4)…` and `B=C_(1)-C_(3)+C_(5)…` then `(B)/(A)=`

Options

A. `"tan"(n pi)/(4)`
B. `"cot"(n pi)/(4)`
C. `"sec"(n pi)/(4)`
D. 1

Answer

To solve the problem, we need to find the ratio \( \frac{B}{A} \) where:

- \( A = C_0 - C_2 + C_4 - C_6 + \ldots \)
- \( B = C_1 - C_3 + C_5 - C_7 + \ldots \)

### Step-by-Step Solution:

1. **Understanding the Binomial Coefficients**:
The coefficients \( C_k \) represent the binomial coefficients from the expansion of \( (1 + x)^n \). Specifically, \( C_k = \binom{n}{k} \).

2. **Using Binomial Expansion**:
We can express \( (1 + x)^n \) using the binomial theorem:
\[ (1 + x)^n = C_0 + C_1 x + C_2 x^2 + C_3 x^3 + C_4 x^4 + \ldots + C_n x^n \]

3. **Substituting \( x = i \) (the imaginary unit)**:
By substituting \( x = i \), we get:
\[ (1 + i)^n = C_0 + C_1 i + C_2 i^2 + C_3 i^3 + C_4 i^4 + \ldots + C_n i^n \]
Here, \( i^2 = -1 \), \( i^3 = -i \), and \( i^4 = 1 \).

4. **Separating Real and Imaginary Parts**:
The real part of \( (1 + i)^n \) will give us \( A \) and the imaginary part will give us \( B \):
\[ A + iB = (1 + i)^n \]

5. **Calculating \( (1 + i)^n \)**:
We can express \( 1 + i \) in polar form:
\[ 1 + i = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) \]
Therefore,
\[ (1 + i)^n = (\sqrt{2})^n \left( \cos \frac{n\pi}{4} + i \sin \frac{n\pi}{4} \right) = 2^{n/2} \left( \cos \frac{n\pi}{4} + i \sin \frac{n\pi}{4} \right) \]

6. **Identifying \( A \) and \( B \)**:
From the above expression, we can identify:
\[ A = 2^{n/2} \cos \frac{n\pi}{4} \]
\[ B = 2^{n/2} \sin \frac{n\pi}{4} \]

7. **Finding the Ratio \( \frac{B}{A} \)**:
Now, we can find the ratio:
\[ \frac{B}{A} = \frac{2^{n/2} \sin \frac{n\pi}{4}}{2^{n/2} \cos \frac{n\pi}{4}} = \frac{\sin \frac{n\pi}{4}}{\cos \frac{n\pi}{4}} = \tan \frac{n\pi}{4} \]

### Final Answer:
\[ \frac{B}{A} = \tan \frac{n\pi}{4} \]