For 0<ϕ<π/2, if x=∑n=0∞ cos2n⁡ϕ,y=∑n=0∞ sin2n⁡ϕ and z=∑n=0∞ cos2n⁡ϕsin2n⁡ϕ then xyz=

x=∏n=0∞ cos2n⁡ϕ=11-cos2θ=cosec2⁡ϕ,y=∑n=0∞ sin2n⁡ϕ=11-sin2θ=sec2⁡ϕz=∑n=0∞ cos2n⁡ϕsin2n⁡ϕ=11−cos2⁡ϕsin2⁡ϕ=11−1xy=xyxy−1.so xyz = xy + z or xyz = x+ y + z as xy = x + y.

For 0<ϕ<π/2, if x=∑n=0∞ cos2n⁡ϕ,y=∑n=0∞ sin2n⁡ϕ and z=∑n=0∞ cos2n⁡ϕsin2n⁡ϕ then xyz=

x=∏n=0∞ cos2n⁡ϕ=11-cos2θ=cosec2⁡ϕ,y=∑n=0∞ sin2n⁡ϕ=11-sin2θ=sec2⁡ϕz=∑n=0∞ cos2n⁡ϕsin2n⁡ϕ=11−cos2⁡ϕsin2⁡ϕ=11−1xy=xyxy−1.so xyz = xy + z or xyz = x+ y + z as xy = x + y.

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

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xy + z

yz + x

xz + y

x + y + z

answer is A, C.

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Detailed Solution

$\begin{array}{l} x = \prod_{n = 0}^{\infty} \cos^{2 n} ϕ = \frac{1}{1 - \cos^{2} θ} = {cosec}^{2} ϕ, y = \sum_{n = 0}^{\infty} \sin^{2 n} ϕ = \frac{1}{1 - \sin^{2} θ} = \sec^{2} ϕ \\ z = \sum_{n = 0}^{\infty} \cos^{2 n} {ϕsin}^{2 n} ϕ = \frac{1}{1 - \cos^{2} {ϕsin}^{2} ϕ} = \end{array}$
$\frac{1}{1 - \frac{1}{xy}} = \frac{xy}{xy - 1}$ .
so xyz = xy + z or xyz = x+ y + z as xy = x + y.