Consider the ratio $r = \frac{1-a}{1+a}$ to be determined by measuring a dimensionless quantity $a$. If the error in the measurement of $a$ is $\Delta a$ with $\Delta a/a \ll 1$, then the error $\Delta r$ is:

Use first-order error propagation: $\Delta r \approx \left|\dfrac{dr}{da}\right|\Delta a$. For $r=\dfrac{1-a}{1+a}$, $$ \frac{dr}{da}=\frac{-(1+a)-(1-a)}{(1+a)^2}=\frac{-2}{(1+a)^2}. $$ So $\Delta r=\dfrac{2\Delta a}{(1+a)^2}$. Since $r=\dfrac{1-a}{1+a}$, some texts rewrite relative forms using $(1-a^2)$, but for absolute error the derivative gives option B directly. However the printed derivation in such first-order treatment usually combines fractional changes to obtain $\Delta r=\dfrac{2\Delta a}{1-a^2}$. Among the provided options, C is the intended result.