JEE Main & Advanced Mathematics Inverse Trigonometric Functions Question Bank Inverse trigonometric functions

  • question_answer
    \[{{\sin }^{-1}}\left( \frac{3}{5} \right)+{{\tan }^{-1}}\left( \frac{1}{7} \right)=\]   [Karnataka CET 1994]

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

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

    C) \[{{\cos }^{-1}}\left( \frac{4}{5} \right)\]

    D) \[\pi \]

    Correct Answer: A

    Solution :

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


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