1.

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

Answer»

x = \(\frac{a-b}{a+b}\) ⇒ \(\frac{1}{x}\) = \(\frac{a+b}{a-b}\)

⇒ \(\frac{1+x}{1-x}\) = \(\frac{a+b+a-b}{a+b-a+b}\) = \(\frac{2a}{2b}\) ⇒ \(\frac{1+x}{1-x}\) = \(\frac{a}{b}\)          (Applying componendo and dividendo)

Similarly, \(\frac{1+y}{1-y}\) = \(\frac{b}{c}\), \(\frac{1+z}{1-z}\) = \(\frac{c}{a}\) ∴ \(\frac{1+x}{1-x}\) . \(\frac{1+y}{1-y}\) . \(\frac{1+z}{1-z}\) = \(\frac{a}{b}\).\(\frac{b}{c}\).\(\frac{c}{a}\) = 1.


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