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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2005

  • question_answer
    If\[0<\phi <\frac{\pi }{2},x=\sum\limits_{n=0}^{\infty }{{{\cos }^{2n}}\phi },y=\sum\limits_{n=0}^{\infty }{{{\sin }^{2n}}\phi }\]and \[z=\sum\limits_{n=0}^{\infty }{{{\cos }^{2n}}\phi }{{\sin }^{2}}n\phi ,\]then:

    A)  \[xyz=xz+y\]                    

    B) \[xyz=xy+z\]

    C)  \[xyz=x+y+z\]

    D)         \[xyz=yz+x\]

    E)  \[xyz=x+yz\]

    Correct Answer: B

    Solution :

    \[\because \]\[x=\sum\limits_{n=0}^{\infty }{{{\cos }^{2n}}\phi =1}+{{\cos }^{2}}\phi +{{\cos }^{4}}\phi +....\] \[=\frac{1}{1-{{\cos }^{2}}\phi }=\frac{1}{{{\sin }^{2}}\phi }\] Similarity\[=\frac{1}{1-{{\sin }^{2}}\phi }=\frac{1}{{{\cos }^{2}}\phi }\] and\[z=\frac{1}{1-{{\sin }^{2}}\phi {{\cos }^{2}}\phi }=\frac{1}{1-\frac{1}{x}.\frac{1}{y}}\]                                 \[=\frac{xy}{xy-1}\] \[\Rightarrow \]               \[xyz-z=xy\] \[\Rightarrow \]               \[xyz=xy+z\]


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