Prove the following

Question:

If $x=\sum_{n=0}^{\infty}(-1)^{n} \tan ^{2 \mathrm{n}} \theta$ and $y=\sum_{n=0}^{\infty} \cos ^{2 n} \theta$, for $0<\theta<\frac{\pi}{4}$

then :

  1. (1) $x(1+y)=1$

  2. (2) $y(1-x)=1$

  3. (3) $y(1+x)=1$

  4. (4) $x(1-y)=1$


Correct Option: , 2

Solution:

$y=1+\cos ^{2} \theta+\cos ^{4} \theta+\ldots . .$

$\Rightarrow y=\frac{1}{1-\cos ^{2} \theta} \Rightarrow \frac{1}{y}=\sin ^{2} \theta$

$x=1-\tan ^{2} \theta+\tan ^{4} \theta+\ldots \ldots$

$x=\frac{1}{1-\left(-\tan ^{2} \theta\right)}=\frac{1}{\sec ^{2} \theta}$

$\Rightarrow x=\cos ^{2} \theta$

$y=\frac{1}{\sin ^{2} \theta} \Rightarrow y=\frac{1}{1-x}$

$\therefore \quad y(1-x)=1$

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