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  3. RAJASTHAN ­ PET

RAJASTHAN ­ PET Rajasthan PET Solved Paper-2011

  • question_answer
    If\[{{\sin }^{-1}}\left( \frac{2a}{1+{{a}^{2}}} \right)+{{\sin }^{-1}}\left( \frac{2b}{1+{{b}^{2}}} \right)=2{{\tan }^{-1}}x,\] then\[x\]is equal to

    A)  \[\frac{a-b}{1+ab}\]

    B)  \[\frac{b}{1+ab}\]

    C)  \[\frac{b}{1-ab}\]

    D)  \[\frac{a+b}{1-ab}\]

    Correct Answer: D

    Solution :

     Use, \[{{\sin }^{-1}}\frac{2x}{1+{{x}^{2}}}=2{{\tan }^{-1}}x\] \[\Rightarrow \] \[2{{\tan }^{-1}}a+2{{\tan }^{-1}}b=2{{\tan }^{-1}}x\] \[\Rightarrow \] \[{{\tan }^{-1}}\left( \frac{a+b}{1-ab} \right)={{\tan }^{-1}}x\] \[\Rightarrow \] \[x=\frac{a+b}{1-ab}\]


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