If `cos(alpha-beta)+cos(beta-gamma)+cos(gamma-alpha)=-3/2` , prove that `cosalpha+cosbeta+cosgamma=sinalpha+sinbeta+singamma=0`

Explanation

To prove that \( \cos \alpha + \cos \beta + \cos \gamma = 0 \) and \( \sin \alpha + \sin \beta + \sin \gamma = 0 \) given that \( \cos(\alpha - \beta) + \cos(\beta - \gamma) + \cos(\gamma - \alpha) = -\frac{3}{2} \), we can follow these steps:

### Step 1: Rewrite the Cosine Terms
Using the cosine of difference identity, we can rewrite the given expression:
\[ \cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \]
\[
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