If `cosx/a=cos(x+theta)/b=cos(x+2theta)/c=cos(x+3theta)/d` then `(a+c)/(b+d)` is equal to `(A)a/d (B)c/d (C)b/c (D)d/a`

Options

A. `(a)/(d)`
B. `(C)/(b)`
C. `(b)/(c)`
D. `(d)/(a)`

Explanation

To solve the problem, we start with the given condition:

\[ \frac{\cos x}{a} = \frac{\cos(x + \theta)}{b} = \frac{\cos(x + 2\theta)}{c} = \frac{\cos(x + 3\theta)}{d} = \frac{1}{k} \]

From this, we can express \(a\), \(b\), \(c\), and \(d\) in terms of \(k\):

1. **Finding expressions for \(a\), \(b\), \(c\), and \(d\)**:
- From \(\frac{\cos x}{a} = \frac{1}{k}\), we have:
\[ a = k \cos x \]
- From \(\frac{\cos(x + \theta)}{b} = \frac{1}{k}\), we have:
\[ b = k \cos(x + \theta) \]
- From \(\frac{\cos(x + 2\theta)}{c} = \frac{1}{k}\), we have:
\[ c = k \cos(x + 2\theta) \]
- From \(\frac{\cos(x + 3\theta)}{d} = \frac{1}{k}\), we have:
\[ d = k \cos(x + 3\theta) \]

2. **Finding \(a + c\) and \(b + d\)**:
- Now we can find \(a + c\):
\[ a + c = k \cos x + k \cos(x + 2\theta) = k (\cos x + \cos(x + 2\theta)) \]
- And for \(b + d\):
\[ b + d = k \cos(x + \theta) + k \cos(x + 3\theta) = k (\cos(x + \theta) + \cos(x + 3\theta)) \]

3. **Using the cosine addition formula**:
- We can use the cosine addition formula:
\[ \cos A + \cos B = 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) \]
- For \(a + c\):
\[ \cos x + \cos(x + 2\theta) = 2 \cos\left(x + \theta\right) \cos\left(\theta\right) \]
- For \(b + d\):
\[ \cos(x + \theta) + \cos(x + 3\theta) = 2 \cos\left(x + 2\theta\right) \cos\left(\theta\right) \]

4. **Substituting back**:
- Now substituting back into our expressions for \(a + c\) and \(b + d\):
\[ a + c = k \cdot 2 \cos(x + \theta) \cos(\theta) \]
\[ b + d = k \cdot 2 \cos(x + 2\theta) \cos(\theta) \]

5. **Finding \(\frac{a + c}{b + d}\)**:
- Now we can find:
\[ \frac{a + c}{b + d} = \frac{k \cdot 2 \cos(x + \theta) \cos(\theta)}{k \cdot 2 \cos(x + 2\theta) \cos(\theta)} \]
- The \(k\) and \(2\cos(\theta)\) cancel out:
\[ \frac{a + c}{b + d} = \frac{\cos(x + \theta)}{\cos(x + 2\theta)} \]

6. **Conclusion**:
- We have shown that:
\[ \frac{a + c}{b + d} = \frac{b}{c} \]
- Therefore, the answer is:
\[ \frac{a + c}{b + d} = \frac{b}{c} \]

Thus, the final answer is \((C) \frac{b}{c}\).