JEE Main & Advanced Mathematics Inverse Trigonometric Functions Question Bank Self Evaluation Test - Inverse Trigonometric Functions

  • question_answer
    If \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{r}^{2}},\] then\[{{\tan }^{-1}}\frac{xy}{zr}+{{\tan }^{-1}}\frac{yz}{xr}+{{\tan }^{-1}}\frac{xz}{yr}=\]

    A) \[\pi \]

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

    C) 0

    D) None of these

    Correct Answer: B

    Solution :

    [b] \[{{\tan }^{-1}}\frac{xy}{zr}+{{\tan }^{-1}}\frac{yz}{xr}+{{\tan }^{-1}}\frac{xz}{yr}\] \[={{\tan }^{-1}}\left[ \frac{\frac{xy}{zr}+\frac{yz}{xr}+\frac{xz}{yr}-\frac{xyz}{{{r}^{3}}}}{1-\left( \frac{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}{{{r}^{2}}} \right)} \right]={{\tan }^{-1}}\infty =\frac{\pi }{2}\]


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