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

  • question_answer
    If \[{{\sin }^{-1}}\frac{1}{3}+{{\sin }^{-1}}\frac{2}{3}={{\sin }^{-1}}x,\]then x is equal to [Roorkee 1995]

    A) 0

    B) \[\frac{\sqrt{5}-4\sqrt{2}}{9}\]

    C) \[\frac{\sqrt{5}+4\sqrt{2}}{9}\]

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

    Correct Answer: C

    Solution :

      \[{{\sin }^{-1}}\frac{1}{3}+{{\sin }^{-1}}\frac{2}{3}\] \[={{\sin }^{-1}}\left[ \frac{1}{3}\sqrt{1-\frac{4}{9}}+\frac{2}{3}\sqrt{1-\frac{1}{9}} \right]={{\sin }^{-1}}\,\left[ \frac{\sqrt{5}+4\sqrt{2}}{9} \right]\] Therefore\[x=\frac{\sqrt{5}+4\sqrt{2}}{9}\].


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