[tex]\left(\frac{1}{a^2}-1\right)\left(\frac{1-a}{1+a}-\frac{1+a}{1-a}\right)\left(1-\frac{a}{4}\right)=\\\\\\\left(\frac{1-a^2}{a^2}\right)\left[\frac{(1-a)^2-(1+a)^2}{(1+a)(1-a)}\right]\left(\frac{4-a}{4}\right)=[/tex]
[tex]\left(\frac{1-a^2}{a^2}\right)\left(\frac{1-2a+a^2-1-2a-a^2}{1-a^2}\right)\left(\frac{4-a}{4}\right)=\\\\\\\frac{1}{a^2}\cdot\frac{-4a}{1}\cdot\frac{4-a}{4}=\\\\\\\frac{-4a(4-a)}{4a^2} = \\\\\\\frac{-1(4-a)}{a}\\\\\\\boxed{\frac{a-4}{a}}[/tex]
