JEE Main & Advanced Sample Paper JEE Main - Mock Test - 5

If \[a=\cos \alpha +i\,\sin \alpha ,\]\[b=\cos \beta +i\sin \beta ,\]\[c=\cos \gamma +isin\gamma \] and \[\frac{b}{c}+\frac{c}{a}+\frac{a}{b}=1,\]then \[\cos \,(\beta -\gamma )+\cos (\gamma -\alpha )+\cos (\alpha -\beta )\] is equal to

A) \[3/2\]

B) \[-3/2\]

C) \[0\]

D) \[1\]

Correct Answer: D

Solution :

\[\frac{b}{c}\,=\frac{\cos \beta +i\sin \beta }{\cos \gamma +i\sin \gamma }\]
Using De'moivre's theorem
\[\frac{b}{c}=\cos (\beta -\gamma )+i\sin (\beta -\gamma )\]	...(i)
Similarly, \[\frac{c}{a}=\cos \,(\gamma -\alpha )+i\sin (\alpha -\beta )\]	...(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, \[\cos (\beta -\gamma )+\cos (\gamma -\alpha )+\cos (\alpha -\beta )=1.\]