A) 2/3
B) 3/2
C) 1/2
D) 1
Correct Answer: B
Solution :
\[\cos \alpha +\cos \beta +\cos \gamma =0\] ??(i) \[\sin \alpha +\sin \beta +\sin \gamma =0\] ?....(ii) Let \[a=\cos \alpha +i\sin \alpha ,\]\[b\,=\cos \beta +i\sin \beta \] \[c=\cos \gamma +i\sin \gamma \] \[\Rightarrow \] \[a+b+c=0\] [by (i) and (ii)] .....(iii) \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\] =\[{{[\cos \alpha +i\sin \alpha ]}^{-1}}+{{[\cos \beta +i\sin \beta ]}^{-1}}+{{[\cos \gamma +i\sin \gamma ]}^{-1}}\] \[=\cos \alpha -i\sin \alpha \]\[+\cos \beta -i\sin \beta \]\[+\cos \gamma -i\sin \gamma \] \[\Rightarrow \,ab+bc+ca=0\] ....(iv) [by (i) and (ii)] Squaring both sides of equation (iii), we get \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca=0\] or \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}=0\] [by (iv)] \[\therefore \,{{[\cos \alpha +i\sin \alpha ]}^{2}}+{{[\cos \beta +i\sin \beta ]}^{2}}+{{[\cos \gamma +i\sin \gamma ]}^{2}}=0\] \[\Rightarrow \,(\cos 2\alpha +\cos 2\beta +\cos 2\gamma )\]\[+i\,(\sin 2\alpha +\sin 2\beta +\sin 2\gamma )=0\] separation of real and imaginary part , \[\Rightarrow \,\,\cos 2\alpha +\cos 2\beta +\cos 2\gamma =0\] .....(v) and \[\sin 2\alpha +\sin 2\beta +\sin 2\gamma \,=0\] .....(vi) \[\Rightarrow \,\,1-2{{\sin }^{2}}\alpha +1-2{{\sin }^{2}}\beta +1-2{{\sin }^{2}}\gamma =0\] [by eq. (v)] \[\therefore \,\,{{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +{{\sin }^{2}}\gamma =\frac{3}{2}\].
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