Question

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

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

Answer 1

The correct option is C

$y\left(1-x\right)=1$

Explanation for the correct option.

Step 1: Solve $x$

$\begin{array}{rcl}x& =& {\left(-1\right)}^0{\tan }^{2\left(0\right)}\theta +{\left(-1\right)}^1{\tan }^{2\left(1\right)}\theta +{\left(-1\right)}^2{\tan }^{2\left(2\right)}\theta +.....\\ & =& 1-{\left(\tan \theta \right)}^2+{\left(\tan \theta \right)}^4+........\end{array}$

From here we can see that, $x$ is in infinite G.P where, $a=1\mathrm{and}r=-{\left(\tan \theta \right)}^2$. So,

$\begin{array}{rcl}x& =& \frac{1}{1+{\tan }^2\theta }\left(\mathrm{sum}\mathrm{of}\mathrm{infinite}\mathrm{G}.\mathrm{P}\right)\\ & =& \frac{1}{se{c}^2\theta }\\ & =& {\cos }^2\theta \end{array}$

Step 2: Solve $y$

$\begin{array}{rcl}y& =& {\cos }^{2\left(0\right)}\theta +{\cos }^{2\left(1\right)}\theta +{\cos }^{2\left(2\right)}\theta +......\\ & =& 1+{\left(\cos \theta \right)}^2+{\left(\cos \theta \right)}^4+......\end{array}$

From here we can see that, $y$ is in infinite G.P where, $a=1\mathrm{and}r={\left(\cos \theta \right)}^2$. So,

$\begin{array}{rcl}y& =& \frac{1}{1-{\cos }^2\theta }\left(\mathrm{sum}\mathrm{of}\mathrm{infinite}\mathrm{G}.\mathrm{P}\right)\\ y& =& \frac{1}{{\sin }^2\theta }\\ \frac{1}{y}& =& {\sin }^2\theta \end{array}$

Step 3: Find $x+\frac{1}{y}$

$\begin{array}{rcl}x+\frac{1}{y}& =& {\cos }^2\theta +{\sin }^2\theta \\ \Rightarrow x+\frac{1}{y}& =& 1\\ \Rightarrow \frac{1}{y}& =& 1-x\\ & \Rightarrow & y\left(1-x\right)=1\end{array}$

Hence, option C is correct.