Question

If $x=\sum_{n=0}^{\infty} \cos ^{2 n} \theta, y=\sum_{n=0}^{\infty} \sin ^{2 n} \theta$
$z=\sum_{n=0}^{\infty} \cos ^{2 n} \theta \sin ^{2 n} \theta$ and $0<\theta<\frac{\pi}{2}$, then

• A. $x z+y z=x y+z$
• B. $x y z=y z+x$
• C. $x y+z=x y+z x$
• D. $x+y+z=x y z+z$

Answer 1

Answer: (A)

$\because x=\displaystyle \sum_{n=0}^{\infty} \cos ^{2 \,n} \theta=\frac{1}{1-\cos ^{2} \theta}$
$=\frac{1}{\sin ^{2} \theta} \left\{\because 0<\theta<\frac{\pi}{2}\right\}$
$y=\displaystyle \sum_{n=0}^{\infty} \sin ^{2\, n} \theta=\frac{1}{1-\sin ^{2} \theta}$
$=\frac{1}{\cos ^{2} \theta}$
$z=\displaystyle \sum_{n=0}^{\infty} \cos ^{2 \,n} \theta \sin ^{2 \,n} \theta$
$=\frac{1}{1-\cos ^{2} \theta \sin ^{2} \theta}$
$\therefore z=\frac{1}{1-\frac{1}{x y}}$
$\Rightarrow x y z-z=x y\,...(i)$

$\because \frac{1}{x}+\frac{1}{y}=\sin ^{2} \theta+\cos ^{2} \theta=1$
$\Rightarrow x+y=x y\,......(ii)$
From Eqs. (i) and (ii),
$ (x+y) z -z =x y$
$ \Rightarrow x z +y z =x y+z $