A) \[\frac{3}{2}\]
B) \[-\frac{3}{2}\]
C) \[0\]
D) 1
Correct Answer: D
Solution :
Given,\[a=cos\text{ }\alpha +i\text{ }sin\text{ }\alpha ,\text{ }b=cos\text{ }\beta +i\text{ }sin\text{ }\beta \] and\[c=cos\text{ }\gamma +2\text{ }sin\,\gamma \] Now, \[\frac{b}{c}=\frac{\cos \beta +i\sin \beta }{\cos \gamma +i\sin \gamma }\times \frac{\cos \gamma -i\sin \gamma }{\cos \gamma -i\sin \gamma }\] \[=\frac{\left[ \begin{align} & \cos \beta \cos \gamma -i\cos \beta \sin \gamma \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+i\sin \beta \cos \gamma -{{i}^{2}}\sin \beta \sin \gamma \\ \end{align} \right]}{{{\cos }^{2}}\gamma +{{\sin }^{2}}\gamma }\] \[=\cos (\beta -\gamma )+i\sin (\beta -\gamma )\] ?.(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) On adding Eqs. (i), (ii) and (iii), \[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=\cos (\beta -\gamma )+\cos (\alpha -\beta )+\cos (\gamma -\alpha )\] \[+i[\sin (\alpha -\beta )+\sin (\beta -\gamma )+\sin (\gamma -\alpha )]=1\] Comparing the real and imaginary parts, \[\cos (\beta -\gamma )+\cos (\alpha -\beta )+\cos (\gamma -\alpha )=1\]
You need to login to perform this action.
You will be redirected in
3 sec
