If \textrm{ }\textrm{ } x = \sum_{n = 0}^{\infty} \left(\left(\right. - 1 \left.\right)\right)^{n} \left(tan\right)^{2 n} \theta \textrm{ }\textrm{ } and \textrm{ }\textrm{ } y = \sum_{n = 0}^{\infty} \left(cos\right)^{2 n} \theta , \textrm{ }\textrm{ } for \textrm{ }\textrm{ } 0 < \theta < \frac{\pi}{4} , \textrm{ }\textrm{ } then :

Engineering

Mathematics

Geometric Progression and Its Properties

Question

$If x = \sum_{n = 0}^{\infty} {(- 1)}^{n} \tan^{2 n} θ and y = \sum_{n = 0}^{\infty} \cos^{2 n} θ, for 0 < θ < \frac{π}{4}, then :$

x(1+y) = 1

x(1–y) = 1

y(1+x) = 1

y(1–x) = 1

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y = 1 + cos²θ + cos⁴θ + .......

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

$x = \frac{1}{1 - (- \tan^{2} θ)} = \frac{1}{\sec^{2} θ}$

⇒ cos²θ = x

$\Rightarrow \frac{1}{y} + x = 1$

⇒ y (1 – x) = 1