If `a=cos alpha+i sin alpha, b=cos beta+isin beta,c=cos gamma+i sin gamma and b/c+c/a+a/b=1,` then `cos(beta-gamma)+cos(gamma-alpha)+cos(alpha-beta)=`

Options

A. `3//2`
B. `-3//2`
C. 0
D. 1

Answer

To solve the problem step by step, we will start by rewriting the given complex numbers in exponential form and then manipulate the equation accordingly.

### Step 1: Rewrite the complex numbers in exponential form
Given:
- \( a = \cos \alpha + i \sin \alpha \)
- \( b = \cos \beta + i \sin \beta \)
- \( c = \cos \gamma + i \sin \gamma \)

Using Euler's formula, we can express these as:
- \( a = e^{i\alpha} \)
- \( b = e^{i\beta} \)
- \( c = e^{i\gamma} \)

### Step 2: Substitute into the given equation
The equation given is:
\[ \frac{b}{c} + \frac{c}{a} + \frac{a}{b} = 1 \]

Substituting the exponential forms:
\[ \frac{e^{i\beta}}{e^{i\gamma}} + \frac{e^{i\gamma}}{e^{i\alpha}} + \frac{e^{i\alpha}}{e^{i\beta}} = 1 \]

This simplifies to:
\[ e^{i(\beta - \gamma)} + e^{i(\gamma - \alpha)} + e^{i(\alpha - \beta)} = 1 \]

### Step 3: Express in terms of cosine and sine
Using Euler's formula again, we can express each term as:
\[ e^{i(\beta - \gamma)} = \cos(\beta - \gamma) + i \sin(\beta - \gamma) \]
\[ e^{i(\gamma - \alpha)} = \cos(\gamma - \alpha) + i \sin(\gamma - \alpha) \]
\[ e^{i(\alpha - \beta)} = \cos(\alpha - \beta) + i \sin(\alpha - \beta) \]

### Step 4: Combine the real and imaginary parts
Now, substituting back into the equation:
\[ \left( \cos(\beta - \gamma) + \cos(\gamma - \alpha) + \cos(\alpha - \beta) \right) + i \left( \sin(\beta - \gamma) + \sin(\gamma - \alpha) + \sin(\alpha - \beta) \right) = 1 \]

### Step 5: Separate real and imaginary parts
For the equation to hold, the real part must equal 1 and the imaginary part must equal 0:
1. Real part:
\[ \cos(\beta - \gamma) + \cos(\gamma - \alpha) + \cos(\alpha - \beta) = 1 \]
2. Imaginary part:
\[ \sin(\beta - \gamma) + \sin(\gamma - \alpha) + \sin(\alpha - \beta) = 0 \]

### Conclusion
Thus, we conclude that:
\[ \cos(\beta - \gamma) + \cos(\gamma - \alpha) + \cos(\alpha - \beta) = 1 \]

### Final Answer
The value of \( \cos(\beta - \gamma) + \cos(\gamma - \alpha) + \cos(\alpha - \beta) \) is:
\[ \boxed{1} \]