The value of $$1-\frac{a}{1-\frac{1}{1+\frac{a}{1-a}}}$$
$$1-\frac{a}{1-\frac{1}{1+\frac{a}{1-a}}}$$
$$=1-\frac{a}{1-\frac{1}{\frac{1-a+a}{1-a}}}$$
$$=1-\frac{a}{1-\frac{1}{\frac{1}{1-a}}}$$
$$=1-\frac{a}{1-(1-a)}$$
$$=1-\frac{a}{a}$$
$$=1-1$$
$$=0$$
Hence, Option D is correct.
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