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VIT Engineering VIT Engineering Solved Paper-2009

  • question_answer
    \[x={{\cos }^{-1}}\left( \frac{1}{\sqrt{1+{{t}^{2}}}} \right),\] \[y={{\sin }^{-1}}\left( \frac{t}{\sqrt{1+{{t}^{2}}}} \right)\Rightarrow \frac{dy}{dx}\]is equal to

    A)  0               

    B)  \[\text{tan}\,t\]

    C)  1                  

    D)  \[\text{sin}\,t\text{ cos}\,t\]

    Correct Answer: C

    Solution :

    Given, \[x={{\cos }^{-1}}\left( \frac{1}{\sqrt{1+{{t}^{2}}}} \right)\] and    \[y={{\sin }^{-1}}\left( \frac{t}{\sqrt{1+{{t}^{2}}}} \right)\] \[\Rightarrow \] \[x={{\tan }^{-1}}t\] and \[y={{\tan }^{-1}}t\] \[\therefore \] \[y=x\] \[\Rightarrow \] \[\frac{dy}{dx}=1\]


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