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

  • question_answer
    If \[{{\sin }^{-1}}x+{{\sin }^{-1}}y+{{\sin }^{-1}}z=\pi ,\] then \[{{x}^{4}}+{{y}^{4}}+{{z}^{4}}+4{{x}^{2}}{{y}^{2}}{{z}^{2}}=k({{x}^{2}}{{y}^{2}}+{{y}^{2}}{{z}^{2}}+{{z}^{2}}{{x}^{2}}).\]where k =

    A) 1

    B) 2

    C) 4

    D) None of these

    Correct Answer: B

    Solution :

    [b] We have, \[{{\sin }^{-1}}x+{{\sin }^{-1}}y=\pi -{{\sin }^{-1}}z\] or \[x\sqrt{(1-{{y}^{2}})}+y\sqrt{(1-{{x}^{2}})}=z\] or \[{{x}^{2}}(1-{{y}^{2}})={{z}^{2}}+{{y}^{2}}(1-{{x}^{2}})-2yz\sqrt{(1-{{x}^{2}})}\] or \[{{({{x}^{2}}-{{z}^{2}}-{{y}^{2}})}^{2}}=4{{y}^{2}}{{z}^{2}}(1-{{x}^{2}})\] or \[{{x}^{4}}+{{y}^{4}}+{{z}^{4}}-2{{x}^{2}}{{z}^{2}}+2{{y}^{2}}{{z}^{2}}-2{{x}^{2}}{{y}^{2}}\] \[+4{{x}^{2}}{{y}^{2}}{{z}^{2}}-4{{y}^{2}}{{z}^{2}}=0\] or \[{{x}^{4}}+{{y}^{4}}+{{z}^{4}}+4{{x}^{2}}{{y}^{2}}{{z}^{2}}=2({{x}^{2}}{{y}^{2}}+{{y}^{2}}{{z}^{2}}+{{z}^{2}}{{x}^{2}})\] \[\therefore k=2\]


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