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BCECE Engineering BCECE Engineering Solved Paper-2012

  • question_answer
    If \[{{\sec }^{-1}}\sqrt{1+{{x}^{2}}}+\cos e{{c}^{-1}}\frac{\sqrt{1+{{y}^{2}}}}{y}\] \[+{{\cot }^{-1}}\frac{1}{z}=\pi ,\]then \[x+y+z\]is equal to

    A)  \[xyz\]                                

    B)         \[2xyz\]                              

    C)         \[xy{{z}^{2}}\]                 

    D)         \[{{x}^{2}}yz\]

    Correct Answer: A

    Solution :

     Given, \[{{\sec }^{-1}}\sqrt{1+{{x}^{2}}}+\cos e{{c}^{-1}}\] \[\frac{\sqrt{1+{{y}^{2}}}}{y}+{{\cot }^{-1}}\frac{1}{z}=\pi \] \[\therefore \]  \[{{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z=\pi \] \[\Rightarrow \]\[{{\tan }^{-1}}\left( \frac{x+y+z-xyz}{1-xy-yz-zx} \right)=\pi \] \[\Rightarrow \]\[x+y+z=xyz\]


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