If 0<ϕ<π2, and x=∑n=0∞ cos2n⁡ ϕ,y=∑n=0∞ sin2n⁡ ϕ z=∑n=0∞ cos2n⁡ ϕsin2n⁡ ϕ,

If 0

A
$x y z = x z + y$
B
$x y z = x z + z$
C
$x y z = x + y + z$
D
$x y z = y z + z$

Solution:

We have $x = \frac{1}{1 - \cos^{2} ϕ} = \frac{1}{\sin^{2} ϕ}$ and $y = \frac{1}{\cos^{2} ϕ}$

$\begin{array}{l} Also, z = \frac{1}{1 - \cos^{2} ϕ \sin^{2} ϕ} = \frac{1}{1 - \frac{1}{x y}} = \frac{x y}{x y - 1} \\ \Rightarrow x y z - z = x y or x y z = x y + z \\ But \frac{1}{x} + \frac{1}{y} = 1 \Rightarrow x + y = x y \\ ∴ \end{array}$

If $0 < ϕ < \frac{π}{2},$ and $x = \sum_{n = 0}^{\infty} \cos^{2 n} ϕ,$
$y = \sum_{n = 0}^{\infty} \sin^{2 n} ϕ$ $z = \sum_{n = 0}^{\infty} \cos^{2 n} ϕ \sin^{2 n} ϕ,$

Solution:

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