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

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

    A) \[\pi \]

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

    C) 0

    D) None of these

    Correct Answer: B

    Solution :

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


You need to login to perform this action.
You will be redirected in 3 sec spinner