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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 8

  • question_answer
    Given that \[0<x<\frac{\pi }{4}\]and \[\frac{\pi }{4}<y<\frac{\pi }{2}\]and \[\sum\limits_{k=0}^{\infty }{{{(-1)}^{k}}{{\tan }^{2k}}x=p;\sum\limits_{k=0}^{\infty }{{{(-1)}^{k}}{{\cot }^{2k}}y=q;}}\]then\[\sum\limits_{k=0}^{\infty }{{{\tan }^{2k}}x{{\cot }^{2k}}y=}\]

    A) \[\frac{1}{p}+\frac{1}{q}-\frac{1}{pq}\]

    B) \[\frac{1}{\left\{ \frac{1}{p}+\frac{1}{q}-\frac{1}{pq} \right\}}\]

    C) \[p+q-pq\]

    D) \[p+q+pq\]

    Correct Answer: B

    Solution :

    [b]: p = Infinite G.P. where \[a=1,r=-{{\tan }^{2}}x\] \[\therefore \]\[p=\frac{a}{1-r}=\frac{1}{1+{{\tan }^{2}}x}={{\cos }^{2}}x\] Similarly, \[q=\frac{1}{1+{{\cot }^{2}}y}={{\sin }^{2}}y\] Now, \[\sum\limits_{k=0}^{\infty }{{{\tan }^{2k}}x{{\cot }^{2k}}}y=\frac{1}{1-{{\tan }^{2}}x{{\cot }^{2}}y}\] \[=\frac{1}{1-\left( \frac{1-{{\cos }^{2}}x}{{{\cos }^{2}}x} \right)\left( \frac{1-{{\sin }^{2}}y}{{{\sin }^{2}}y} \right)}\] \[=\frac{pq}{p+q-1}=\frac{1}{\left\{ \frac{1}{p}+\frac{1}{q}-\frac{1}{pq} \right\}}\]


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