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Manipal Engineering Manipal 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) \[\tan t\]

    C)  1                            

    D)        \[\sin t\cos t\]

    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        \[y={{\tan }^{-1}}t\]                 \[y=x\]   \[\Rightarrow \]   \[\frac{dy}{dx}=1\]


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