Question

For $0<\varphi <\pi /2$ if $x={\sum }_{n=0}^{\infty }{\cos }^{2n}\varphi ,y={\sum }_{n=0}^{\infty }{\sin }^{2n}\varphi ,z={\sum }_{n=0}^{\infty }{\cos }^{2n}\varphi {\sin }^{2n}\varphi$, then

• A. $xyz=xz+y$
• B. $xyz=xy+z$
• C. $xyz=x+y+z$
• D. $xyz=yz+x$

Answer 1

Answer: (B)

$xyz=xy+z$

Given $x=\sum _{n=0}^{\infty }{\cos }^{2n}\varphi ,y=\sum _{n=0}^{\infty }{\sin }^{2n}\varphi$ and $z=\sum _{n=0}^{\infty }{\cos }^{2n}\varphi {\sin }^{2n}\varphi$

\since $0<\varphi <\frac{\pi }{2}$, so each series is geometric series with common ratio $r<1$. Therefore, the series are convergent.

Now, $x=\frac{1}{1-{\cos }^2\varphi }$

$=\frac{1}{{\sin }^2\varphi }$ $(\because S_{\infty }=\frac{a}{1-r})$

$y=\frac{1}{1-{\sin }^2\varphi }$ $(\because S_{\infty }=\frac{a}{1-r})$

$=\frac{1}{{\cos }^2\varphi }$

$z=\frac{1}{1-{\sin }^2{\varphi \cos }^2\varphi }$

$(\because S_{\infty }=\frac{a}{1-r})$

Consider, $xyz=\frac{1}{{\sin }^2\varphi {\cos }^2\varphi (1-{\sin }^2{\varphi \cos }^2\varphi )}$ $\left(1\right)$

Also, $=\frac{1}{{\sin }^2\varphi {\cos }^2\varphi }+\frac{1}{1-{\sin }^2{\varphi \cos }^2\varphi }$

$xy+z=\frac{1-{\sin }^2{\varphi \cos }^2\varphi +{\sin }^2\varphi {\cos }^2\varphi }{{\sin }^2\varphi {\cos }^2\varphi (1-{\sin }^2{\varphi \cos }^2\varphi )}$

$=\frac{1}{{\sin }^2\varphi {\cos }^2\varphi (1-{\sin }^2{\varphi \cos }^2\varphi )}$

$=xyz$ $\left[From\right(1\left)\right]$