If 0 < θ , ϕ < π 2 , x = ∑ n = 0 ∞ cos 2 n θ , y = ∑ n = 0 ∞ sin 2 n ϕ and z = ∑ n = 0 ∞ cos 2 n θ · sin 2 n ϕ then :

x = 1 1 - cos 2 θ ⇒ sin 2 θ = 1 x Also, cos 2 θ = 1 y & 1 - sin 2 θ cos 2 θ = 1 z So, 1 - 1 x × 1 y = 1 z ⇒ z x y - 1 = x y ...... i Also, 1 x + 1 y = 1 ⇒ x + y = x y ...... ii From i and ii x y + z = x y z = x + y z

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Mathematics - Sequence & Series Question with Solution | TestHub

MathematicsSequence & SeriesMiscellaneous/MixedHard2 minPYQ_2021

MathematicsHardstatement

If $0 < θ, ϕ < \frac{π}{2}, x = \sum_{n = 0}^{\infty} \cos^{2 n} θ, y = \sum_{n = 0}^{\infty} \sin^{2 n} ϕ$ and $z = \sum_{n = 0}^{\infty} \cos^{2 n} θ \cdot \sin^{2 n} ϕ$ then :

Options:

Answer:

Solution:

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

Also, $\cos^{2} θ = \frac{1}{y} & 1 - \sin^{2} θ \cos^{2} θ = \frac{1}{z}$

So, $1 - \frac{1}{x} \times \frac{1}{y} = \frac{1}{z} \Rightarrow z (x y - 1) = x y$ ...... $(i)$

Also, $\frac{1}{x} + \frac{1}{y} = 1 \Rightarrow x + y = x y$ ...... $(ii)$

From $(i)$ and $(ii)$

$x y + z = x y z = (x + y) z$

Stream:JEESubject:MathematicsTopic:Sequence & SeriesSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2021

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