Question

If $x=\sum _{n=0}^{\infty }(-1{)}^n{\tan }^{2n}\theta$ and $y=\sum _{n=0}^{\infty }{\cos }^{2n}\theta ,$ where $0<\theta <\frac{\pi }{4},$ then:

• A. $y(1+x)=1$
• B. $x(1-y)=1$
• C. $y(1-x)=1$
• D. $x(1+y)=1$

Answer 1

The correct option is C $y(1-x)=1$

$y=1+{\cos }^2\theta +{\cos }^4\theta +\cdots$
$\Rightarrow y=\frac{1}{1-{\cos }^2\theta }\Rightarrow \frac{1}{y}={\sin }^2\theta$
$x=1-{\tan }^2\theta +{\tan }^4\theta -\cdots$
$\Rightarrow x=\frac{1}{1-(-{\tan }^2\theta )}={\cos }^2\theta$
$\therefore x+\frac{1}{y}=1\Rightarrow y(1-x)=1$