A) 1
B) \[-1\]
C) \[\frac{\sqrt{3}}{2}\]
D) \[\frac{1}{\sqrt{2}}\]
Correct Answer: B
Solution :
\[\sin {{600}^{o}}\cos {{330}^{o}}+\cos {{120}^{o}}\sin {{150}^{o}}\] \[=\sin ({{720}^{o}}-{{120}^{o}})\cos ({{180}^{o}}+{{150}^{o}})\] \[+\cos {{120}^{o}}\sin {{150}^{o}}\] \[=\sin {{120}^{o}}\cos {{150}^{o}}+\cos {{120}^{o}}\sin {{150}^{o}}\] \[=\sin ({{120}^{o}}+{{150}^{o}})\] \[=\sin {{270}^{o}}=\sin ({{180}^{o}}+{{90}^{o}})\] \[=-\sin {{90}^{o}}=-1\]
You need to login to perform this action.
You will be redirected in
3 sec
