Question

If $0<\theta ,\varphi <\frac{\mathrm{\pi }}{2},x=\sum _{n=0}^{\infty }{\cos }^{2n}\theta ,y=\sum _{n=0}^{\infty }{\sin }^{2n}\varphi ,$and $z=\sum _{n=0}^{\infty }{\cos }^{2n}\theta ·{\sin }^{2n}\varphi$ then:

• A. $xyz=4$
• B. $xy-z=\left(x+y\right)z$
• C. $xy+yz+zx=z$
• D. $xy+z=\left(x+y\right)z$

Answer 1

The correct option is D

$xy+z=\left(x+y\right)z$

Explanation for the correct option.

Find the correct relation between $x,y,$and $z$.

The sum of an infinite geometric series with first term $a$ and common ratio $r$ is given as: $S_{\infty }=\frac{a}{1-r}$.

Now, $x=\sum _{n=0}^{\infty }{\cos }^{2n}\theta$ is an infinte geometric series whose first term is $0$ and the common ratio is ${\cos }^2\theta$. So it can be written as:

$$\begin{gathered} x=\sum _{n=0}^{\infty }{\cos }^{2n}\theta \\ \Rightarrow x=1+{\cos }^2\theta +{\cos }^4\theta +...\infty \\ \Rightarrow x=\frac{1}{1-{\cos }^2\theta } \\ \Rightarrow \left(1-{\cos }^2\theta \right)=\frac{1}{x} \\ \Rightarrow 1-\frac{1}{x}={\cos }^2\theta ...\left(1\right) \end{gathered}$$

Now, $y=\sum _{n=0}^{\infty }{\sin }^{2n}\varphi$ is an infinte geometric series whose first term is $0$ and the common ratio is ${\sin }^2\varphi$. So it can be written as:

$$\begin{gathered} y=\sum _{n=0}^{\infty }{\sin }^{2n}\varphi \\ \Rightarrow y=1+{\sin }^2\varphi +{\sin }^4\varphi +...\infty \\ \Rightarrow y=\frac{1}{1-{\sin }^2\varphi } \\ \Rightarrow \left(1-{\sin }^2\varphi \right)=\frac{1}{y} \\ \Rightarrow 1-\frac{1}{y}={\sin }^2\varphi ...\left(2\right) \end{gathered}$$

Now, $z=\sum _{n=0}^{\infty }{\cos }^{2n}\theta ·{\sin }^{2n}\varphi$ is an infinte geometric series whose first term is $0$ and the common ratio is ${\cos }^2\theta {\sin }^2\varphi$. So it can be written as:

$$\begin{gathered} z=\sum _{n=0}^{\infty }{\cos }^{2n}\theta ·{\sin }^{2n}\varphi \\ \Rightarrow z=1+{\cos }^2\theta ·{\sin }^2\varphi +{\cos }^4\theta ·{\sin }^4\varphi +...\infty \\ \Rightarrow z=\frac{1}{1-{\cos }^2\theta ·{\sin }^2\varphi } \\ \Rightarrow z=\frac{1}{1-\left(1-{\displaystyle \frac{1}{x}}\right)\left(1-{\displaystyle \frac{1}{y}}\right)}\left[\mathrm{Using}\mathrm{equations}1\mathrm{and}2\right] \\ \Rightarrow z=\frac{1}{1-\left({\displaystyle \frac{x-1}{x}}\right)\left({\displaystyle \frac{y-1}{y}}\right)} \\ \Rightarrow z=\frac{xy}{xy-\left(x-1\right)\left(y-1\right)} \end{gathered}$$

Now cross multiply and find the relation.

$$\begin{gathered} xyz-\left(x-1\right)\left(y-1\right)z=xy \\ \Rightarrow xyz-\left(xy-x-y+1\right)z=xy \\ \Rightarrow xyz-xyz+xz+yz-z=xy \\ \Rightarrow xz+yz-z=xy \\ \Rightarrow (x+y)z=xy+z \end{gathered}$$

So, the relation between $x,y,$and $z$ is $xy+z=\left(x+y\right)z$.

Hence, the correct option is D.