A) \[sin\text{ }2(\alpha +\beta +\gamma )\]
B) \[2sin(\alpha +\beta +\gamma )\]
C) 1
D) 0
Correct Answer: D
Solution :
Let\[a=\cos \alpha +i\sin \alpha ,b=\cos \beta +i\sin \beta \]and\[c=\cos \gamma +i\sin \gamma ,\] \[\therefore \]\[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=(\cos \alpha -i\sin \alpha )+(\cos \beta -\sin \beta )\] \[+(\cos \gamma -i\sin \gamma )\] \[\Rightarrow \]\[\frac{bc+ac+ab}{abc}=(\cos \alpha +\cos \beta +\cos \gamma )\] \[-i(\sin \alpha +\sin \beta +\sin \gamma )\] \[\left[ \begin{align} & \because given\cos \alpha +\cos \beta +\cos \gamma =0 \\ & and\,\,\,\,\,\,\,\,\,\,\,\sin \alpha +\sin \beta +\sin \gamma =0 \\ \end{align} \right]\] \[\Rightarrow \] \[\frac{ab+bc+ca}{abc}=0-i0\] \[\Rightarrow \] \[ab+bc+ca=0\] \[abc=(\cos \alpha +i\sin \alpha )+(\cos \beta +i\sin \beta )\] \[+(\cos \gamma +i\sin \gamma )\] \[=(\cos \alpha +\cos \beta +\cos \gamma )+i(\sin \alpha +\sin \beta +\sin \gamma )\] \[=0+i0=0\] \[\therefore \]\[{{(a+b+c)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2(ab+bc+ca)=0\] \[\Rightarrow \] \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}=0\] \[\Rightarrow \]\[{{(\cos \alpha +i\sin \alpha )}^{2}}+{{(\cos \beta +i\sin \beta )}^{2}}\] \[+{{(\cos \gamma +i\sin \gamma )}^{2}}=0\] \[\Rightarrow \]\[(\cos 2\alpha +i\sin 2\alpha )+(\cos 2\beta +i\sin 2\beta )\] \[+(\cos 2\gamma +i\sin 2\gamma )=0+i0\] (from De-moiver's theorem) \[\Rightarrow \] \[(\cos 2\alpha +\cos 2\beta +\cos 2\gamma )\] \[+i(\sin 2\alpha +\sin 2\beta +\sin 2\gamma )=0+0i\] \[\therefore \] \[cos2\alpha +\cos 2\beta +\cos 2\gamma =0\] and \[sin2\alpha +sin2\beta +sin2\gamma =0\]
You need to login to perform this action.
You will be redirected in
3 sec
