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

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

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

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

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

$$\ \ \ \frac{\ 1}{1-a}+\ \frac{\ 1}{1-b}+\ \frac{\ 1}{1-c}-3=1$$

$$\ \ \ \frac{\ 1}{1-a}+\ \frac{\ 1}{1-b}+\ \frac{\ 1}{1-c}=4$$