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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 43

  • question_answer
    If \[\tan \left( \frac{\pi }{4}\left( {{n}^{2}}+2n \right) \right)=\cot \left( \frac{\pi }{4}\left( {{n}^{2}}+n+1 \right) \right),\]\[n\in I,\] then

    A) n should be an odd integer   

    B) \[n=4k-1,\,k\in I\]           

    C) \[n=3k,\,k\in I\]                           

    D) None of these                                            

    Correct Answer: A

    Solution :

    [a] \[\tan \left( \frac{\pi }{4}({{n}^{2}}+2n) \right)=\tan \left( \frac{\pi }{4}(2-({{n}^{2}}+n+1) \right)\] \[=\frac{\pi }{4}\left( {{n}^{2}}+2n \right)=k\pi +\frac{\pi }{4}\left( 2-\left( {{n}^{2}}+n+1 \right) \right),\,\,k,\,n\in I\] \[\Rightarrow \,\,\,\,\,\,{{n}^{2}}+2n=4k+(1-{{n}^{2}}-n)\] \[\Rightarrow \,\,\,\,2{{n}^{2}}+3n=4k+1\] \[\Rightarrow \,\,\,\,n=4\lambda +1,\,\lambda \in I\]                       


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