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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2002

  • question_answer
    If\[y={{\tan }^{-1}}\left( \frac{\cos x-\sin x}{\cos x+\sin x} \right),\]then\[\frac{dy}{dx}\]is equal to

    A)  1

    B)  \[-1\]

    C)  \[\sin x\]

    D)  \[\cos x\]

    Correct Answer: B

    Solution :

     \[y={{\tan }^{-1}}\left( \frac{1-\tan x}{1+\tan x} \right)={{\tan }^{-1}}\tan \left( \frac{\pi }{4}-x \right)\] \[\therefore \] \[y=\frac{\pi }{4}-x\] \[\Rightarrow \] \[\frac{dy}{dx}=-1\]


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