A) \[\frac{6}{65}\]
B) \[\frac{3}{\sqrt{130}}\]
C) \[-\frac{3}{\sqrt{130}}\]
D) \[-\frac{3}{65}\]
Correct Answer: B
Solution :
\[\therefore \]\[{{(\sin \alpha +\sin \beta )}^{2}}+{{(\cos \alpha +\cos \beta )}^{2}}\] \[={{\left( -\frac{21}{65} \right)}^{2}}+{{\left( -\frac{27}{65} \right)}^{2}}\] \[\Rightarrow \] \[2+2\cos (\alpha -\beta )=\frac{1170}{4225}\] \[\Rightarrow \] \[4{{\cos }^{2}}\left( \frac{\alpha -\beta }{2} \right)=\frac{1170}{4225}\] \[\Rightarrow \]\[{{\cos }^{2}}\left( \frac{\alpha -\beta }{2} \right)=\frac{1170}{4\times {{(65)}^{2}}}=\frac{1170}{{{(130)}^{2}}}=\frac{9}{130}\] \[\Rightarrow \]\[\cos \left( \frac{\alpha -\beta }{2} \right)=\frac{3}{\sqrt{130}}\]
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