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Q. 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:

KEAMKEAM 2005

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 $