Manipal Engineering Manipal Engineering Solved Paper-2010

  • question_answer
    The value of\[{{\sin }^{-1}}\left[ \cos \left( {{\sin }^{-1}}\sqrt{\frac{2-\sqrt{3}}{4}} \right. \right.\]\[\left. \left. +{{\cos }^{-1}}\frac{\sqrt{12}}{4}+{{\sec }^{-1}}\sqrt{2} \right) \right]\]is

    A)  0                                            

    B) \[\frac{\pi }{4}\]

    C) \[\frac{\pi }{6}\]                              

    D)        \[\frac{\pi }{2}\]

    Correct Answer: A

    Solution :

    We have, \[{{\sin }^{-1}}\left[ \cos \left( {{\sin }^{-1}}\sqrt{\left( \frac{2-\sqrt{3}}{4} \right)}+{{\cos }^{-1}}\frac{\sqrt{12}}{4}+{{\sec }^{-1}}\sqrt{2} \right) \right]\] \[={{\sin }^{-1}}\left[ \cos \left( {{\sin }^{-1}}\left( \frac{\sqrt{3}-1}{2\sqrt{2}} \right)+{{\cos }^{-1}}\frac{\sqrt{3}}{2}+{{\cos }^{-1}}\frac{1}{\sqrt{2}} \right) \right]\]                                 \[\left[ \because \,\,{{\left( \frac{2-\sqrt{3}}{4} \right)}^{1/2}}=\frac{\sqrt{3}-1}{2\sqrt{2}} \right]\]                 \[={{\sin }^{-1}}[\cos ({{15}^{o}}+{{30}^{o}}+{{45}^{o}})]\]                 \[={{\sin }^{-1}}[\cos {{90}^{o}}]\]                            \[\left( \because \,\,\sin {{15}^{o}}=\frac{\sqrt{3}-1}{2\sqrt{2}} \right)\]                 \[={{\sin }^{-1}}0\]                 \[=0\]


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