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

  • question_answer
    What is the value of \[\tan \left( {{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z \right)-\cot (co{{t}^{-1}}x+co{{t}^{-1}}y+co{{t}^{-1}}z)?\]

    A) 0

    B) \[2(x+y+z)\]

    C) \[\frac{3\pi }{2}\]

    D) \[\frac{3\pi }{2}+x+y+z\]  

    Correct Answer: A

    Solution :

    [a] \[\tan (ta{{n}^{-1}}x+ta{{n}^{-1}}y+ta{{n}^{-1}}z)\] \[-\cot (co{{t}^{-1}}x+co{{t}^{-1}}y+co{{t}^{-1}}z)\] \[=\tan (ta{{n}^{-1}}x+ta{{n}^{-1}}y+ta{{n}^{-1}}z)\] \[-\cot \left( \frac{\pi }{2}-{{\tan }^{-1}}x+\frac{\pi }{2}-{{\tan }^{-1}}y+\frac{\pi }{2}-{{\tan }^{-1}}z \right)\] \[\left( \because {{\tan }^{-1}}x+{{\cot }^{-1}}x=\frac{\pi }{2} \right)\] \[=\tan (ta{{n}^{-1}}x+ta{{n}^{-1}}y+ta{{n}^{-1}}z)\] \[-\cot \{3\pi /2-(ta{{n}^{-1}}x+ta{{n}^{-1}}y+ta{{n}^{-1}}z)\] \[=\tan (ta{{n}^{-1}}x+ta{{n}^{-1}}y+ta{{n}^{-1}}z)\] \[-\tan (ta{{n}^{-1}}x+ta{{n}^{-1}}y+ta{{n}^{-1}}z)=0\]


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