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

  • question_answer
    \[{{\cos }^{-1}}\left( \frac{15}{17} \right)+2{{\tan }^{-1}}\left( \frac{1}{5} \right)=\]   [EAMCET 1981]

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

    B) \[{{\cos }^{-1}}\left( \frac{171}{221} \right)\]

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

    D) None of these

    Correct Answer: D

    Solution :

      \[{{\cos }^{-1}}\left( \frac{15}{17} \right)+2{{\tan }^{-1}}\left( \frac{1}{5} \right)\]\[={{\cos }^{-1}}\left( \frac{15}{17} \right)+{{\cos }^{-1}}\left( \frac{1-1/25}{1+1/25} \right)\] \[={{\cos }^{-1}}\left( \frac{15}{17} \right)+{{\cos }^{-1}}\left( \frac{12}{13} \right)\] \[={{\cos }^{-1}}\left( \frac{15}{17}\times \frac{12}{13}-\sqrt{1-{{\left( \frac{15}{17} \right)}^{2}}}\sqrt{1-{{\left( \frac{12}{13} \right)}^{2}}} \right)\] \[={{\cos }^{-1}}\left( \frac{140}{221} \right)\].


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