For 0 2 , if x = n = 0 cos 2 n , y = n = 0 sin 2 n and z = n = 0 cos 2 n ? sin 2 n , then

Question

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

Smart Achievers · Accepted Answer

(3)

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

and $z = \frac{1}{1 - \sin^{2} ϕ \cdot \cos^{2} ϕ}$

Clearly, $\frac{1}{x} + \frac{1}{y} = 1$

$\Rightarrow$ $x y = x + y$ and $\frac{1}{z} = 1 - \frac{1}{x y}$

$\Rightarrow$ $x y = x y z - z$

$∴$ $x y z = x y + z = x + y + z$

Solution

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

1 $x y z = x z + y$

2 $x y z = x y - z$

3 $x y z = x + y + z$

4 $x y z = y z + x$

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Solution

Q. For 0<ϕ<π2, if x=∑n=0∞cos2nϕ, y=∑n=0∞sin2nϕ and z=∑n=0∞cos2nϕ sin2nϕ, then

1 xyz=xz+y

2 xyz=xy-z

3 xyz=x+y+z

4 xyz=yz+x