If $$0$$<θ,ϕ<π$$/2$$ and $$x=$$∑$$n=0$$∞ $$sin2n$$⁡ϕ, $$y=$$∑$$n=0$$∞ $$cos2n$$⁡θ and $$z=$$∑$$n=0$$∞ cosn⁡$$($$θ$$+$$ϕ$$)cosn$$⁡$$($$θ−ϕ$$)$$, then

Correct option is 3xyz-xy=yz-zx

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

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detailed solution

Correct option is C

We have x=11−sin2⁡ϕ=1cos2⁡ϕ,y=11−cos2⁡θ=1sin2⁡θ,Also, cos⁡(θ+ϕ)cos⁡(θ−ϕ)=cos2⁡ϕ−sin2⁡θ=1x−1yAs, 0

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If 0<θ,ϕ<π/2 and x=∑n=0∞ sin2n⁡ϕ, y=∑n=0∞ cos2n⁡θ and z=∑n=0∞ cosn⁡(θ+ϕ)cosn⁡(θ−ϕ), then

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