• # question_answer If $a=\cos \alpha +i\,\sin \alpha ,\,\,b=\cos \beta +i\,\sin \beta ,$$c=\cos \gamma +i\,\sin \gamma \,\,\text{and}\,\,\frac{b}{c}+\frac{c}{a}+\frac{a}{b}=1,$ then $\cos (\beta -\gamma )+\cos (\gamma -\alpha )+\cos (\alpha -\beta )$ is equal to [RPET 2001] A) 3/2 B) - 3/2 C) 0 D) 1

$\frac{b}{c}=\frac{\cos \beta +i\sin \beta }{\cos \gamma +i\sin \gamma }\times \frac{\cos \gamma -i\sin \gamma }{\cos \gamma -i\sin \gamma }$ $\Rightarrow \frac{b}{c}=\,\cos (\beta -\gamma )+i\sin (\beta -\gamma )$              ......(i) Similarly,$\,\frac{c}{a}=\cos (\gamma -\alpha )+i\sin \,(\gamma -\alpha )$    ......(ii)  and    $\,\,\frac{a}{b}=\cos (\alpha -\beta )+i\sin (\alpha -\beta )$  .....(iii) from (i) + (ii) + (iii) $\cos (\beta -\gamma )+\cos (\gamma -\alpha )+\cos (\alpha -\beta )$$+i[\sin (\beta -\gamma )+\sin (\gamma -\alpha )+\sin (\alpha -\beta )]=1$ Equating real and imaginary parts, $\cos (\beta -\gamma )+\cos (\gamma -\alpha )+\cos (\alpha -\beta )=1$.
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