If x- n-0 - a n -y- n-0 - b n -z- n-0 - ab n-where a-b-1 then prove that xz-yz-xy-z-Another form-For 0- - - 2 if x- n-0 -cos 2n - y- n-0 -sin 2n - - z- n-0 -cos 2n -sin 2n - then prove that xz-yz-z-xy-

x= 1 1 a ⇒ a= x 1 x Similarly b= y 1 y , ab= z 1 z ∴ z 1 z = x 1 x = y 1 y or xyz xy=z(xy x y+1) ∴ xz+yz=xy+z ...

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Byju's Answer

Standard XII

Mathematics

Property 7

If x=∑ n=0 ...

Question

If $x = \infty \sum n = 0 a^{n}, y = \infty \sum n = 0 b^{n}, z = \infty \sum n = 0 {(a b)}^{n}$ ,
where $a, b < 1$ then prove that $x z + y z = x y + z$ .
Another form:
For $0 < θ, ϕ - \frac{π}{2}$ if $x = \infty \sum n = 0 {cos}^{2 n} θ$
$y = \infty \sum n = 0 {sin}^{2 n} ϕ, z = \infty \sum n = 0 {cos}^{2 n} θ {sin}^{2 n} ϕ$ then prove that $x z + y z - z = x y$ .

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Solution

$x = \frac{1}{1 - a}$ $\Rightarrow a = \frac{x - 1}{x}$
Similarly $b = \frac{y - 1}{y}, a b = \frac{z - 1}{z}$
$∴ \frac{z - 1}{z} = \frac{x - 1}{x} = \frac{y - 1}{y}$
or $x y z - x y = z (x y - x - y + 1)$
$∴ x z + y z = x y + z$

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Similar questions

Q. For

0 < ϕ < \frac{π}{2}

, if

x = \sum_{n = 0}^{\infty} {cos}^{2 n} ϕ, y = \sum_{n = 0}^{\infty} {sin}^{2 n} ϕ

and

z = \sum_{n = 0}^{\infty} {cos}^{2 n} ϕ {sin}^{2 n} ϕ

, then show that

x y z = x y + z

Q. If

0 < θ < \frac{π}{2}

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x = \sum_{n = 0}^{\infty} {cos}^{2 n} θ, y = \sum_{n = 0}^{\infty} {sin}^{2 n} θ, z = \sum_{n = 0}^{\infty} {cos}^{2 n} θ {sin}^{2 n} θ

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x y z

For 0< $ϕ$ < $\frac{π}{2}, i f x = \sum_{n = 0}^{\infty} c o s^{2 n} ϕ, y = \sum_{n = 0}^{\infty} s i n^{2 n} ϕ, z = \sum_{n = 0}^{\infty} c o s^{2 n} ϕ, t h e n$

Q. If

0 < θ, ϕ < \frac{π}{2},

x = \infty \sum n = 0 {cos}^{2 n} θ,

y = \infty \sum n = 0 {sin}^{2 n} ϕ

and

z = \infty \sum n = 0 {cos}^{2 n} θ \cdot {sin}^{2 n} ϕ

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Q. For

0 < ϕ < π / 2

x = \sum_{n = 0}^{\infty} {cos}^{2 n} ϕ, y = \sum_{n = 0}^{\infty} {sin}^{2 n} ϕ, z = \sum_{n = 0}^{\infty} {cos}^{2 n} ϕ {sin}^{2 n} ϕ

, then

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If x=∞∑n=0an,y=∞∑n=0bn,z=∞∑n=0(ab)n,where a,b<1 then prove that xz+yz=xy+z.Another form:For 0<θ,ϕ−π2 if x=∞∑n=0cos2nθy=∞∑n=0sin2nϕ,z=∞∑n=0cos2nθsin2nϕ then prove that xz+yz−z=xy.

x=11−a ⇒ a=x−1xSimilarly b=y−1y,ab=z−1z∴z−1z=x−1x=y−1yor xyz−xy=z(xy−x−y+1)∴xz+yz=xy+z

$x = \frac{1}{1 - a}$ $\Rightarrow a = \frac{x - 1}{x}$
Similarly $b = \frac{y - 1}{y}, a b = \frac{z - 1}{z}$
$∴ \frac{z - 1}{z} = \frac{x - 1}{x} = \frac{y - 1}{y}$
or $x y z - x y = z (x y - x - y + 1)$
$∴ x z + y z = x y + z$