Question

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

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

Answer 1

The correct option is D $xy+z=(x+y)z$

$x=1+{\cos }^2\theta +{\cos }^4\theta +\cdots \infty$
$\Rightarrow x=\frac{1}{1-{\cos }^2\theta }(\text{For GP}:S_{\infty }=\frac{a}{1-r})\Rightarrow x=\frac{1}{{\sin }^2\theta }\dots \left(1\right)$

$y=1+{\sin }^2\varphi +{\sin }^4\varphi +\cdots \infty$
$\Rightarrow y=\frac{1}{1-{\sin }^2\varphi }\Rightarrow y=\frac{1}{{\cos }^2\varphi }\dots \left(2\right)$

$z=\frac{1}{1-{\cos }^2\theta \cdot {\sin }^2\varphi }=\frac{1}{1-(1-\frac{1}{x})(1-\frac{1}{y})}\left[\text{From (1) and (2)}\right]z=\frac{xy}{xy-(x-1)(y-1)}$

$\Rightarrow xz+yz-z=xy$
$\Rightarrow xy+z=(x+y)z$