BCECE Engineering BCECE Engineering Solved Paper-2002

  • question_answer
    \[{{\sin }^{-1}}\frac{3}{5}+{{\tan }^{-1}}\frac{1}{7}\]is equal to:

    A) \[\pi /2\]                             

    B)  \[{{\cos }^{-1}}4/5\]      

    C)         \[\pi \]                

    D)         \[\pi /4\]

    Correct Answer: D

    Solution :

    \[{{\sin }^{-1}}\frac{3}{5}+{{\tan }^{-1}}\frac{1}{7}\] \[={{\tan }^{-1}}\frac{3}{4}+{{\tan }^{-1}}\frac{1}{7}\] \[={{\tan }^{-1}}\left( \frac{\frac{3}{4}+\frac{1}{7}}{1-\frac{3}{4}.\frac{1}{7}} \right)\] \[={{\tan }^{-1}}\left( \frac{25}{25} \right)={{\tan }^{-1}}(1)\] \[=\frac{\pi }{4}\]


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