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JEE Main & Advanced Mathematics Inverse Trigonometric Functions Question Bank Critical Thinking

  • question_answer
    If\[{{\cos }^{-1}}p+{{\cos }^{-1}}q+{{\cos }^{-1}}r=\pi \]then\[{{p}^{2}}+{{q}^{2}}+{{r}^{2}}+2pqr=\] [Karnataka CET 2004]

    A) 3

    B) 1

    C) 2

    D) -1

    Correct Answer: B

    Solution :

     Trick: According to given condition, we put   \[p=q=r=\frac{1}{2}\]. Then,\[{{p}^{2}}+{{q}^{2}}+{{r}^{2}}+2pqr\] = \[{{\left( \frac{1}{2} \right)}^{2}}+{{\left( \frac{1}{2} \right)}^{2}}+{{\left( \frac{1}{2} \right)}^{2}}+2.\frac{1}{2}.\frac{1}{2}.\frac{1}{2}\]             = \[\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{2}{8}\]=1.


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