If `cosalpha+cosbeta+cosgamma=0=sinalpha+sinbeta+singamma`, then which of the following is/are true:- (a)`cos(alpha-beta)+cos(beta-gamma)+cos(gamma-delta)=-3/2` (b)`cos(alpha-beta)+cos(beta-gamma)+cos(gamma-delta)=-1/2` (c)`sumcos2alpha+2cos(alpha+beta)+2cos(beta+gamma)+2cos(gamma+alpha)=0` (d)`sumsin2alpha+2sin(alpha+beta)+2sin(beta+gamma)+2sin(gamma+alpha)=0`

Options

A. A is true and B is false
B. A is false and B is true
C. Both A and B are true
D. Both A and B are false

Answer

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

1. \( \cos \alpha + \cos \beta + \cos \gamma = 0 \)
2. \( \sin \alpha + \sin \beta + \sin \gamma = 0 \)

From these equations, we can derive several relationships and analyze the options provided.

### Step 1: Analyze the equations

We can rewrite the first equation as:
\[ \cos \alpha + \cos \beta = -\cos \gamma \]
And the second equation as:
\[ \sin \alpha + \sin \beta = -\sin \gamma \]

### Step 2: Square both sides

Squaring both sides of the first equation gives:
\[ (\cos \alpha + \cos \beta)^2 = \cos^2 \gamma \]
Expanding the left side:
\[ \cos^2 \alpha + \cos^2 \beta + 2 \cos \alpha \cos \beta = \cos^2 \gamma \]

### Step 3: Rearranging the equation

Rearranging gives:
\[ 2 \cos \alpha \cos \beta = \cos^2 \gamma - \cos^2 \alpha - \cos^2 \beta \]
This is our first derived equation.

### Step 4: Similar approach for sine

Using the second equation, we can square it as well:
\[ (\sin \alpha + \sin \beta)^2 = \sin^2 \gamma \]
Expanding gives:
\[ \sin^2 \alpha + \sin^2 \beta + 2 \sin \alpha \sin \beta = \sin^2 \gamma \]

Rearranging gives:
\[ 2 \sin \alpha \sin \beta = \sin^2 \gamma - \sin^2 \alpha - \sin^2 \beta \]
This is our second derived equation.

### Step 5: Adding the two derived equations

Now, we can add the two derived equations:
\[ 2 \cos \alpha \cos \beta + 2 \sin \alpha \sin \beta = (\cos^2 \gamma - \cos^2 \alpha - \cos^2 \beta) + (\sin^2 \gamma - \sin^2 \alpha - \sin^2 \beta) \]

### Step 6: Simplifying using Pythagorean identity

Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can simplify:
\[ \cos^2 \gamma + \sin^2 \gamma = 1 \]
Thus, we can rewrite the right side as:
\[ 1 - ( \cos^2 \alpha + \sin^2 \alpha + \cos^2 \beta + \sin^2 \beta ) = 1 - 2 \]
This gives us:
\[ 2 \cos \alpha \cos \beta + 2 \sin \alpha \sin \beta = -1 \]

### Step 7: Final simplification

This can be rewritten as:
\[ \cos(\alpha - \beta) = -\frac{1}{2} \]
This means:
\[ \cos(\alpha - \beta) + \cos(\beta - \gamma) + \cos(\gamma - \alpha) = -\frac{3}{2} \]

### Conclusion

From the analysis, we find that option (a) is true:
- (a) \( \cos(\alpha - \beta) + \cos(\beta - \gamma) + \cos(\gamma - \alpha) = -\frac{3}{2} \)