A) \[0\]
B) \[1\]
C) \[2\]
D) \[3\]
Correct Answer: B
Solution :
Since,\[A+B+C={{270}^{o}}\] \[\therefore \] \[A=B=C={{90}^{o}}\] \[\therefore \] \[\cos 2A+\cos 2B+\cos 2C\] \[+\,\,4\sin A\sin B\sin C\] \[=\cos {{180}^{o}}+\cos {{180}^{o}}+\cos {{180}^{o}}+\] \[4\sin {{90}^{o}}\sin {{90}^{o}}\sin {{90}^{o}}\] \[=-1-1-1+4\] \[=1\]You need to login to perform this action.
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