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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2001

  • question_answer
    The value of \[\left[ {{\tan }^{-1}}\left( \frac{1}{2} \right)+{{\tan }^{-1}}\left( \frac{1}{3} \right) \right]\]is:

    A) \[\frac{\pi }{2}\]                                              

    B) \[\pi \]

    C)  \[\pi /4\]                            

    D)  none of these

    Correct Answer: C

    Solution :

     \[=\left[ {{\tan }^{-1}}\frac{1}{2}+{{\tan }^{-1}}\frac{1}{3} \right],\] \[{{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \frac{x+y}{1-xy} \right)\] Therefore \[={{\tan }^{-1}}\left[ \frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}.\frac{1}{3}} \right]\Rightarrow {{\tan }^{-1}}\left[ \frac{5/6}{5/6} \right]\] \[\Rightarrow \]               \[{{\tan }^{-1}}1\] \[=\pi /4\]


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