Test Index

Sequences and Series Part 2

Section: Mathematics

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Question : 96 of 100

Marks: +1, -0

If

0 < \theta, \phi < \frac{\pi}{2}, x=\sum\limits_{n=0}^{\infty} \cos^{2n} \theta, y=\sum\limits_{n=0}^{\infty} \sin^{2n} \phi

z=\sum\limits_{n=0}^{\infty} \cos^{2n} \theta \cdot \sin^{2n} \phi

[25 Feb 2021 Shift 1]

$x y-z=(x+y) z$
$x y+y z+z x=z$
$x y z=4$
$x y+z=(x+y) z$

Solution:

Given,

x=\sum\limits_{n=0}^{\infty} \cos^{2n} \theta

\;y\;\; =\sum\limits_{n=0}^{\infty} \sin^{2n} \phi

\;z=\sum\limits_{n=0}^{\infty} \cos^{2n} \theta \cdot \sin^{2n} \phi

\Rightarrow\;

x\;\; =1+\cos^{2} \theta+\cos^{4} \theta+\dots \infty

\;x=\frac{1}{1-\cos^{2} \theta}

. . . (i)

\;1-\cos^{2} \theta=\frac{1}{x} \Rightarrow \cos^{2} \theta=1-\frac{1}{x}

\Rightarrow\;y=1+\sin^{2} \phi+\sin^{4} \phi+\dots \infty

\therefore \;y=\frac{1}{1-\sin^{2} \phi}

. . . (ii)

\;1-\sin^{2} \phi=\frac{1}{y} \Rightarrow \sin^{2} \phi=1-\frac{1}{y}

\Rightarrow\;z=1+\cos^{2} \theta \cdot \sin^{2} \phi+\cos^{4} \theta \sin^{4} \phi+\dots \infty

\therefore \;z=\frac{1}{1-\cos^{2} \theta \sin^{2} \phi}

. . . (iii)

From Eqs. (i), (ii) and (iii), we get

\;z=\frac{1}{1-\left(1-\frac{1}{x}\right)\left(1-\frac{1}{y}\right)}

\left[\begin{array}{l} \because \cos^{2} \theta = 1-\frac{1}{x} \\ \because \sin^{2} \phi = 1-\frac{1}{y} \end{array}\right]

\;z=\frac{x y}{x y-(x-1)(y-1)}

\;z=\frac{x y}{x y-x y+x+y-1}

\Rightarrow\;\; x z+y z-z=x y

\Rightarrow\;\; x y+z=(x+y) z

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