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

# If $0<\theta ,\varphi <\pi /2$ and $x=\sum _{n=0}^{\mathrm{\infty }} {\mathrm{sin}}^{2n}\varphi ,$ $y=\sum _{n=0}^{\mathrm{\infty }} {\mathrm{cos}}^{2n}\theta$ and $z=\sum _{n=0}^{\mathrm{\infty }} {\mathrm{cos}}^{n}\left(\theta +\varphi \right){\mathrm{cos}}^{n}\left(\theta -\varphi \right)$, then

1. A

$xyz+1=yz-zx$

2. B

$xyz-1=yz+zx$

3. C

$xyz-xy=yz-zx$

4. D

$xyz+1=yz+zx$

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### Solution:

We have $x=\frac{1}{1-{\mathrm{sin}}^{2}\varphi }=\frac{1}{{\mathrm{cos}}^{2}\varphi },$
$y=\frac{1}{1-{\mathrm{cos}}^{2}\theta }=\frac{1}{{\mathrm{sin}}^{2}\theta },$

Also, $\mathrm{cos}\left(\theta +\varphi \right)\mathrm{cos}\left(\theta -\varphi \right)={\mathrm{cos}}^{2}\varphi -{\mathrm{sin}}^{2}\theta =\frac{1}{x}-\frac{1}{y}$
As, $0<{\mathrm{cos}}^{2}\varphi <1,0<{\mathrm{sin}}^{2}\theta <1,-1<{\mathrm{cos}}^{2}\varphi -{\mathrm{sin}}^{2}\theta <1$

Thus, $z=\sum _{n=0}^{\mathrm{\infty }} {\left(\frac{1}{x}-\frac{1}{y}\right)}^{n}=\frac{1}{1-\left(\frac{1}{x}-\frac{1}{y}\right)}=\frac{xy}{xy-y+x}$
or $xyz-xy=yz-xz$  