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TS EAMCET 4 May 2018 Shift 1 Solved Paper

Section: Mathematics

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Question : 22 of 160

Marks: +1, -0

If x =

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

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

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

\frac{\pi}{2}

, then

xz + yz = xy + z
xyz = yz + x
xy + z = xy + zx
x + y + z = xyz + z

Solution:

It is given that,

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

m

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

And,

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

\text{So},

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

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

=\frac{1}{\sin^{2}\theta}

Also,

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

=\frac{1}{1-\sin^{2}\theta}

=\frac{1}{\cos^{2}\theta}

Therefore,

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

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

=\frac{1}{1-\frac{1}{x y}}

x y z - z = x y

…… (I)

Also,

\frac{1}{x} + \frac{1}{y} = \sin^{2}\theta + \cos^{2}\theta

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

x + y = x y

Therefore, equation (I) can be written as,

(x + y) z - z = x y

x z + y z = x y + z

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