If 0<θ, ϕ<π2, x=∑n=0∞cos2nθ,y=∑n=0∞sin2nϕ and z=∑n=0∞cos2nθsin2nϕ then

x=1+cos2θ+cos4θ+...... in G.Px=a1−r=11−cos2θ In G.P. S∝ =a1−r1−1x=cos2θSimilarly 1−1y=sin2ϕand z=11−cos2θsin2ϕ=11−(1−1x)(1−1y)z=xyxy−(x−1)(y−1)By solving xy+z=(x+y)z

If 0<θ, ϕ<π2, x=∑n=0∞cos2nθ,y=∑n=0∞sin2nϕ and z=∑n=0∞cos2nθsin2nϕ then

x=1+cos2θ+cos4θ+...... in G.Px=a1−r=11−cos2θ In G.P. S∝ =a1−r1−1x=cos2θSimilarly 1−1y=sin2ϕand z=11−cos2θsin2ϕ=11−(1−1x)(1−1y)z=xyxy−(x−1)(y−1)By solving xy+z=(x+y)z

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

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$xyz = 4$

$xy - z = (x + y) z$

$x y + y z + z x = z$

$xy + z = (x + y) z$

answer is D.

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Detailed Solution

$x = 1 + \cos^{2} θ + \cos^{4} θ + . ..... in G . P$
$x = \frac{a}{1 - r} = \frac{1}{1 - \cos^{2} θ} (In G . P . S_{\propto}^{} = \frac{a}{1 - r})$
$1 - \frac{1}{x} = \cos^{2} θ$
Similarly $1 - \frac{1}{y} = \sin^{2} ϕ$
and $z = \frac{1}{1 - \cos^{2} {θsin}^{2} ϕ} = \frac{1}{1 - (1 - \frac{1}{x}) (1 - \frac{1}{y})}$
$z = \frac{xy}{xy - (x - 1) (y - 1)}$
By solving $xy + z = (x + y) z$