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

∵0<ϕ<π2 ∴ 0<sin⁡ϕ<1 and 0<cos⁡ϕ<1 ∴ x=∑n=0∞ cos2n⁡ϕ=1+cos2⁡ϕ+cos4⁡ϕ+…+∞ =11−cos2⁡ϕ=1sin2⁡ϕ or sin2⁡ϕ=1x ...(i) and y=∑n=0∞ sin2n⁡ϕ=1+sin2⁡ϕ+sin4⁡ϕ+…+∞ =11−sin2⁡ϕ=1cos2⁡ϕ or cos2⁡ϕ=1y ....(ii)From Eqs. (i) and (ii),sin2ϕ+cos2ϕ=1x+1y 1=1x+1y∴ xy=x+y ....(iii)and z=∑n=0∞cos2nϕsin2nϕ=1+cos2⁡ϕsin2⁡ϕ+cos4⁡ϕsin4⁡ϕ+…=11−sin2⁡ϕcos2⁡ϕ=11−1xy[ from Eqs. (i) and (ii)] ⇒ z=xyxy−1⇒ xyz=z+xy and xyz=z+x+y from Eq.(iii)

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

∵0<ϕ<π2 ∴ 0<sin⁡ϕ<1 and 0<cos⁡ϕ<1 ∴ x=∑n=0∞ cos2n⁡ϕ=1+cos2⁡ϕ+cos4⁡ϕ+…+∞ =11−cos2⁡ϕ=1sin2⁡ϕ or sin2⁡ϕ=1x ...(i) and y=∑n=0∞ sin2n⁡ϕ=1+sin2⁡ϕ+sin4⁡ϕ+…+∞ =11−sin2⁡ϕ=1cos2⁡ϕ or cos2⁡ϕ=1y ....(ii)From Eqs. (i) and (ii),sin2ϕ+cos2ϕ=1x+1y 1=1x+1y∴ xy=x+y ....(iii)and z=∑n=0∞cos2nϕsin2nϕ=1+cos2⁡ϕsin2⁡ϕ+cos4⁡ϕsin4⁡ϕ+…=11−sin2⁡ϕcos2⁡ϕ=11−1xy[ from Eqs. (i) and (ii)] ⇒ z=xyxy−1⇒ xyz=z+xy and xyz=z+x+y from Eq.(iii)

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

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$z + y x = z y x$

$xyz = xz + y$

$xyz = x + y + z$

$xyz = yz + x$

answer is B, C.

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Detailed Solution

$∵ 0 < ϕ < \frac{π}{2} ∴ 0 < \sin ϕ < 1 and 0 < \cos ϕ < 1 ∴ x = \sum_{n = 0}^{\infty} \cos^{2 n} ϕ = 1 + \cos^{2} ϕ + \cos^{4} ϕ + \dots + \infty = \frac{1}{1 - \cos^{2} ϕ} = \frac{1}{\sin^{2} ϕ} or \sin^{2} ϕ = \frac{1}{x} . . . (i) and y = \sum_{n = 0}^{\infty} \sin^{2 n} ϕ = 1 + \sin^{2} ϕ + \sin^{4} ϕ + \dots + \infty = \frac{1}{1 - \sin^{2} ϕ} = \frac{1}{\cos^{2} ϕ} or \cos^{2} ϕ = \frac{1}{y} . . . . (ii)$
From Eqs. (i) and (ii),
$\sin^{2} ϕ + \cos^{2} ϕ = \frac{1}{x} + \frac{1}{y} 1 = \frac{1}{x} + \frac{1}{y}$
$∴ xy = x + y . . . . (iii)$
and $z = \sum_{n = 0}^{\infty} \cos^{2 n} {ϕsin}^{2 n} ϕ$
$\begin{array}{l} = 1 + \cos^{2} {ϕsin}^{2} ϕ + \cos^{4} {ϕsin}^{4} ϕ + \dots \\ = \frac{1}{1 - \sin^{2} {ϕcos}^{2} ϕ} = \frac{1}{1 - \frac{1}{xy}} [from Eqs. (i) and (ii)] \end{array}$
$\begin{array}{lrl} \Rightarrow z = \frac{xy}{xy - 1} \\ \Rightarrow xyz = z + xy \\ and xyz = z + x + y [from Eq . (iii)] \end{array}$