A) \[-\frac{3}{\sqrt{130}}\]
B) \[\frac{3}{\sqrt{130}}\]
C) \[\frac{6}{65}\]
D) \[-\frac{6}{65}\]
Correct Answer: A
Solution :
Given that, \[\sin \alpha +\sin \beta =-\frac{21}{65}\] ...(i) and \[\cos \alpha +\cos \beta =-\frac{27}{65}\] ...(ii) Squaring Eqs. (i) and (ii) and then adding, we get \[{{(\sin \alpha +\sin \beta )}^{2}}+{{(\cos \alpha +\cos \beta )}^{2}}\] \[={{\left( -\frac{21}{65} \right)}^{2}}+{{\left( -\frac{27}{65} \right)}^{2}}\] \[{{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +2\sin \alpha \sin \beta +{{\cos }^{2}}\alpha +{{\cos }^{2}}\beta \] \[+2\cos \alpha \cos \beta =\frac{1170}{4225}\] \[\Rightarrow \] \[2+2(\cos \alpha \cos \beta +\sin \alpha \sin \beta )=\frac{1170}{4225}\] \[\Rightarrow \] \[2+2\cos (\alpha -\beta )=\frac{1170}{4225}\] \[\Rightarrow \] \[2[1+\cos (\alpha -\beta )]=\frac{1170}{4225}\] \[\Rightarrow \] \[2\left[ 2{{\cos }^{2}}\left( \frac{\alpha -\beta }{2} \right) \right]=\frac{1170}{4225}\] \[\Rightarrow \] \[{{\cos }^{2}}\left( \frac{\alpha -\beta }{2} \right)=\frac{1170}{4\times 4225}\] \[\Rightarrow \] \[{{\cos }^{2}}\left( \frac{\alpha -\beta }{2} \right)=\frac{9}{130}\] \[\Rightarrow \] \[\cos \left( \frac{\alpha -\beta }{2} \right)=-\frac{3}{\sqrt{130}}\] \[(\because \pi <\alpha -\beta <3\pi )\]
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