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

A
$x y z + 1 = y z - z x$
B
$x y z - 1 = y z + z x$
C
$x y z - x y = y z - z x$
D
$x y z + 1 = y z + z x$

Solution:

We have $x = \frac{1}{1 - \sin^{2} ϕ} = \frac{1}{\cos^{2} ϕ},$
$y = \frac{1}{1 - \cos^{2} θ} = \frac{1}{\sin^{2} θ},$

Also, $\cos (θ + ϕ) \cos (θ - ϕ) = \cos^{2} ϕ - \sin^{2} θ = \frac{1}{x} - \frac{1}{y}$
As, $0 < \cos^{2} ϕ < 1, 0 < \sin^{2} θ < 1, - 1 < \cos^{2} ϕ - \sin^{2} θ < 1$
$\Rightarrow - 1 < \frac{1}{x} - \frac{1}{y} < 1$
Thus, $z = \sum_{n = 0}^{\infty} {(\frac{1}{x} - \frac{1}{y})}^{n} = \frac{1}{1 - (\frac{1}{x} - \frac{1}{y})} = \frac{x y}{x y - y + x}$
$\Rightarrow z (x y - y + x) = x y$ or $x y z - x y = y z - x z$

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

Solution:

Related content