A) 0
B) 1
C) -1
D) None of the above
Correct Answer: B
Solution :
Given, \[a=\cos \alpha +i\sin \alpha ,b=\cos \beta +i\sin \beta ,\] \[c=\cos \gamma +i\sin \gamma \] and \[\frac{b}{c}+\frac{c}{a}+\frac{a}{b}=1\] \[\Rightarrow \] \[\frac{\cos \beta +i\sin \beta }{\cos \gamma +i\sin \gamma }+\frac{\cos \gamma +i\sin \gamma }{\cos \alpha +i\sin \alpha }\] \[+\frac{\cos \alpha +i\sin \alpha }{\cos \beta +i\sin \beta }=1\] \[\Rightarrow \] \[\frac{{{e}^{i\beta }}}{{{e}^{i\gamma }}}+\frac{{{e}^{i\gamma }}}{{{e}^{i\alpha }}}+\frac{{{e}^{i\alpha }}}{{{e}^{i\beta }}}=1\] \[\Rightarrow \] \[{{e}^{i(\beta +\gamma )}}+{{e}^{i(\gamma -\alpha )}}+{{e}^{i(\alpha -\beta )}}=1\] \[\Rightarrow \]\[\cos (\beta -\gamma )+i\sin (\beta -\gamma )+\cos (\gamma -\alpha )\] \[+i\sin (\gamma -\alpha )+\cos (\alpha -\beta )+i\sin (\alpha -\beta )=1\] \[\Rightarrow \]\[[\cos (\beta -\gamma )+\cos (\gamma -\alpha )+\cos (\alpha -\beta )]\] \[+i[\sin (\beta -\gamma )+\sin (\gamma -\alpha )+\sin (\alpha -\beta )]\] \[=1+i0\] On comparing real parts both sides \[\cos (\beta -\gamma )+\cos (\gamma -\alpha )+\cos (\alpha -\beta )=1\]
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