A) -1
B) 1
C) \[\frac{1}{\sqrt{2}}\]
D) \[\frac{\sqrt{3}}{2}\]
Correct Answer: A
Solution :
\[\sin \,{{600}^{o}}\,\cos \,{{330}^{o}}+\cos \,{{120}^{o}}\,\sin \,{{150}^{o}}\] \[=-\sin \,{{60}^{o}}\,\cos \,{{30}^{o}}-\sin \,{{30}^{o}}\,\cos \,{{60}^{o}}\] \[=-\left\{ \sin \,({{60}^{o}}+{{30}^{o}}) \right\}=-1\].
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