If x-a-ba-b- y-b-cb-c and z-c-ac-a- find the value of 1-x1-y1-z1-x1-y1-z-

The correct option is B 1 x=a ba+b #160; #160; #160;....GivenSo, by componendo dividendo ⇒x+1x 1 = a b+(a+b)a b (a+b) = 2a2b = ab ...

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Standard VIII

Mathematics

Division of a Polynomial by a Polynomial

If x=a-ba+b...

Question

If $x = \frac{a - b}{a + b}, y = \frac{b - c}{b + c}$ and $z = \frac{c - a}{c + a}$ , find the value of $\frac{(1 + x) (1 + y) (1 + z)}{(1 - x) (1 - y) (1 - z)}$ .

a b c

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a^{2} b^{2} c^{2}

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Solution

The correct option is B $1$
$x = \frac{a - b}{a + b}$ ....Given
So, by componendo-dividendo
$\Rightarrow \frac{x + 1}{x - 1} = \frac{a - b + (a + b)}{a - b - (a + b)} = - \frac{2 a}{2 b} = - \frac{a}{b}$
$∴ \frac{x + 1}{1 - x} = \frac{a}{b}$ ....(1)
Similarly, $\frac{y + 1}{1 - y} = \frac{b}{c}$ and $\frac{z + 1}{1 - z} = \frac{c}{a}$ ...(2)

$∴ \frac{(1 + x) (1 + y) (1 + z)}{(1 - x) (1 - y) (1 - z)} = \frac{a}{b} \times \frac{b}{c} \times \frac{c}{a} = 1$ ....From (1) and (2)

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If x=a−ba+b,y=b−cb+c and z=c−ac+a, find the value of (1+x)(1+y)(1+z)(1−x)(1−y)(1−z).

The correct option is B 1x=a−ba+b ....GivenSo, by componendo-dividendo⇒x+1x−1=a−b+(a+b)a−b−(a+b)=−2a2b=−ab∴x+11−x=ab ....(1)Similarly, y+11−y=bc and z+11−z=ca ...(2)∴(1+x)(1+y)(1+z)(1−x)(1−y)(1−z)=ab×bc×ca=1 ....From (1) and (2)

If $x = \frac{a - b}{a + b}, y = \frac{b - c}{b + c}$ and $z = \frac{c - a}{c + a}$ , find the value of $\frac{(1 + x) (1 + y) (1 + z)}{(1 - x) (1 - y) (1 - z)}$ .