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JCECE Engineering JCECE Engineering Solved Paper-2004

  • question_answer
    If\[A+B+C={{270}^{o}}\], then \[\cos 2A+\cos 2B\]\[+\cos 2C+4\sin A\sin B\sin C\]is equal to:

    A) \[0\]                                     

    B) \[1\]

    C) \[2\]                                     

    D) \[3\]

    Correct Answer: B

    Solution :

    Since,\[A+B+C={{270}^{o}}\]      \[\therefore \]      \[A=B=C={{90}^{o}}\] \[\therefore \]       \[\cos 2A+\cos 2B+\cos 2C\]                 \[+\,\,4\sin A\sin B\sin C\]                 \[=\cos {{180}^{o}}+\cos {{180}^{o}}+\cos {{180}^{o}}+\] \[4\sin {{90}^{o}}\sin {{90}^{o}}\sin {{90}^{o}}\]                 \[=-1-1-1+4\]                 \[=1\]


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