VIT Engineering VIT Engineering Solved Paper-2009

  • question_answer
    \[{{\cos }^{-1}}\left( \frac{-1}{2} \right)-2{{\sin }^{-1}}\left( \frac{1}{2} \right)+3{{\cos }^{-1}}\left( \frac{-1}{\sqrt{2}} \right)\] \[-4{{\tan }^{-1}}(-1)\]equals

    A)  \[\frac{19\pi }{12}\]

    B)  \[\frac{35\pi }{12}\]

    C)  \[\frac{47\pi }{12}\]

    D)  \[\frac{43\pi }{12}\]

    Correct Answer: D

    Solution :

    \[{{\cos }^{-1}}\left( -\frac{1}{2} \right)-2{{\sin }^{-1}}\left( \frac{1}{2} \right)+3{{\cos }^{-1}}\left( -\frac{1}{\sqrt{2}} \right)\] \[-4\,{{\tan }^{-1}}(-1)\] \[=\pi -{{\cos }^{-1}}\left( \frac{1}{2} \right)-2\left( \frac{\pi }{6} \right)+3\left( \pi -{{\cos }^{-1}}\left( \frac{1}{\sqrt{2}} \right) \right)\] \[+\,4{{\tan }^{-1}}(1)\] \[=\pi -\frac{\pi }{3}-\frac{\pi }{3}+3\left( \pi -\frac{\pi }{4} \right)+4\cdot \frac{\pi }{4}\] \[=\frac{\pi }{3}+3\cdot \frac{3\pi }{4}+\pi \] \[=\frac{43\pi }{12}\]


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