Question 122

If $$\frac{a}{1-a} + \frac{b}{1-b} + \frac{c}{1-c} = 1$$ the the value of $$\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c}$$

Solution

Expression : $$\frac{a}{1-a} + \frac{b}{1-b} + \frac{c}{1-c} = 1$$

Let's put each term equal to each other

=> $$3\frac{a}{1 - a} = 1$$

=> $$3a = 1 - a$$

=> $$a = \frac{1}{4} = b = c$$

To find : $$\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c}$$

= $$\frac{1}{1 - \frac{1}{4}} + \frac{1}{1 - \frac{1}{4}} + \frac{1}{1 - \frac{1}{4}}$$

= $$3 \times \frac{4}{3} = 4$$

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