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

The correct option is D 1 Given, #160; x=a ba+b, y = b cb+c, z=c ac+a Thus 1+x=1+a ba+b=2aa+b Similarly, 1+y=2bb+c, 1+z=2cc+a and 1 ...

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Standard XII

Mathematics

Properties of Inequalities

If x=a-ba+b...

Question

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

a b c

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

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1

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Solution

The correct option is D $1$
Given, $x = \frac{a - b}{a + b}, y = \frac{b - c}{b + c}, z = \frac{c - a}{c + a}$
Thus $1 + x = 1 + \frac{a - b}{a + b} = \frac{2 a}{a + b}$
Similarly, $1 + y = \frac{2 b}{b + c}, 1 + z = \frac{2 c}{c + a}$
and $1 - x = \frac{2 b}{a + b}, 1 - y = \frac{2 c}{b + c}, 1 - z = \frac{2 a}{c + a}$
Therefore, $\frac{(1 + x) (1 + y) (1 + z)}{(1 - x) (1 - y) (1 - z)} = \frac{(2 a) (2 b) (2 c)}{(2 b) (2 c) (2 a)} = 1$

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

The correct option is D 1Given, x=a−ba+b,y=b−cb+c,z=c−ac+aThus 1+x=1+a−ba+b=2aa+bSimilarly, 1+y=2bb+c,1+z=2cc+aand 1−x=2ba+b,1−y=2cb+c,1−z=2ac+aTherefore, (1+x)(1+y)(1+z)(1−x)(1−y)(1−z)=(2a)(2b)(2c)(2b)(2c)(2a)=1

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