A) \[\frac{1}{\sqrt{2}}\]
B) \[\frac{1}{2}\]
C) \[\frac{\sqrt{3}}{2}\]
D) 1
Correct Answer: B
Solution :
Given that \[sin\left( 60{}^\circ -x \right)=cos(y+60{}^\circ )\] \[\Rightarrow \] \[\cos \left( {{90}^{{}^\circ }}-{{60}^{{}^\circ }}+x \right)=\cos \left( y+{{60}^{{}^\circ }} \right)\] \[\Rightarrow \]\[30{}^\circ +x=y+60{}^\circ \] \[\Rightarrow \] \[x-y=60{}^\circ -30{}^\circ \Rightarrow x-y=30{}^\circ \] \[\therefore sin\text{ }\left( xy \right)=sin30{}^\circ =\]You need to login to perform this action.
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