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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2002

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

    A) \[-1\]                                    

    B) \[\sin 2\pi \]

    C)  \[\cos 2x\]                        

    D)  zero

    Correct Answer: A

    Solution :

    \[y={{\tan }^{-1}}\left( \frac{\cos x-\sin x}{\cos x+\sin x} \right)\] \[y={{\tan }^{-1}}\left( \frac{1-\tan x}{1+\tan x} \right)\] \[\Rightarrow \]               \[y={{\tan }^{-1}}\tan \left( \frac{\pi }{4}-2 \right)\] \[y=\frac{\pi }{4}-x\Rightarrow \frac{dy}{dx}=-1\]


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